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*Symbolic Logic: Syntax, Semantics, and Proof*

7.3 Logical Properties

7.3 Logical Properties

**Branch**: “A branch includes all the propositions obtained by starting from the bottom of the tree and reading upward through the tree” (109).

**Fully decomposed branch**: “A branch is fully decomposed when all propositions in the branch that can be decomposed have been decomposed” (110).

**Partially decomposed branch**: “A branch is partially decomposed when there is at least one proposition in the branch that has not been decomposed” (110).

**Closed branch**: “A closed branch contains a proposition ‘

**P**’ and its literal negation ‘

**¬P**.’ A closed branch is represented by an ‘

**X**’ ” (111).

**Open branch**: “An open branch is a branch that is not closed, that is, a branch that does not contain a proposition ‘

**P**’ and its literal negation ‘¬

**P**’ ” (111).

**Completed open branch**: “A completed open branch is a fully decomposed branch that is not closed. That is, it is a fully decomposed branch that does not contain a proposition and its literal negation. An open branch is denoted by writing an ‘

**0**’ at the bottom of the tree” (112).

**Completed open tree**: “A tree is a completed open tree if and only if it has at least one completed open branch. That is, a tree is a completed open tree if and only if it contains at least one fully decomposed branch that is not closed. A completed open tree is a tree where there is at least one branch that has an ‘

**0**’ under it” (113).

**Closed tree**: “A tree is closed when all of the tree’s branches are closed. A closed tree will have an ‘

**X**’ under every branch” (114).] In order to do this, Agler will first redefine what a completed open branch is in RL, and he will explain how “analyzing trees in RL is different from analyzing trees in PL” (Agler 294). Agler defines completed open branch, closed tree, and closed branch in the following ways for RL.

To illustrate what makes a completed open branch, Agler first shows one that looks like it could be an example but in fact is not.Completed open branch: A branch is a completed open branch if and only if (1) all complex propositions that can be decomposed into atomic propositions or negated atomic propositions are decomposed; (2) for all universally quantified propositions ‘(∀x)P’ occurring in the branch, there is a substitution instance ‘P(a/x)’ for each constant that occurs in that branch; and (3) the branch is not a closed branch.Closed tree: A tree is a closed tree if and only if all branches close.Closed branch: A branch is a closed branch if and only if there is a proposition and its literal negation (e.g., ‘P’ and ‘¬P’).(Agler 294)

1 | (∀x)(Px→Qx) | P |

2 | (∃x)(Px∧¬Qx)✔ | P |

3 | Pa→Qa✔ | 1∀D |

4 | Pb∧¬Qb✔ | 2∃D |

5 | Pb | 4∧D |

6 | ¬Qb / \ / \ | 4∧D |

7 | ¬Pa Qa | 3→D |

For all universally quantified propositions ‘(∀x)P’ occurring in the branch, there is a substitution instance ‘P(a/x)’ for each constant that occurs in that branch.(294d)

1 | (∀x)(Px→Qx) | P |

2 | (∃x)(Px∧¬Qx)✔ | P |

3 | Pa→Qa✔ | 1∀D |

4 | Pb∧¬Qb✔ | 2∃D |

5 | Pb | 4∧D |

6 | ¬Qb / \ / \ | 4∧D |

7 | ¬Pa Qa | 3→D |

8 | Pb→Qb Pb→Qb / \ / \ | 1∀D |

9 | ¬Pb Qb ¬Pb Qb X X X X | 8→D |

(∀x)(¬Px→¬Rx), (∀x)(Rx→Px)(Agler 295)

1 | (∀x)(¬Px→¬Rx) | P |

2 | (∀x)(Rx→Px) | P |

1 | (∀x)(¬Px→¬Rx) | P |

2 | (∀x)(Rx→Px) | P |

3 | ¬Pa→¬Ra | 1∀D |

1 | (∀x)(¬Px→¬Rx) | P |

2 | (∀x)(Rx→Px) | P |

3 | ¬Pa→¬Ra | 1∀D |

4 | Ra→Pa | 2∀D |

1 | (∀x)(¬Px→¬Rx) | P |

2 | (∀x)(Rx→Px) | P |

3 | ¬Pa→¬Ra✔ | 1∀D |

4 | Ra→Pa✔ / \ | 2∀D |

5 | ¬¬Pa ¬Ra / \ / \ | 3→D |

6 | ¬Ra Pa ¬Ra Pa O O O O | 4→D |

**Closed branch**: A branch is a closed branch if and only if there is a proposition and its literal negation (e.g., ‘

**P**’ and ‘¬

**P**’);” “

**Completed open branch**: A branch is a completed open branch if and only if (1) all complex propositions that can be decomposed into atomic propositions or negated atomic propositions are decomposed; (2) for all universally quantified propositions ‘(∀x)

**P**’ occurring in the branch, there is a substitution instance ‘

**P**(a/

*x*)’ for each constant that occurs in that branch; and (3) the branch is not a closed branch” (Agler 294). As we can see, we have fulfilled the requirements for a completed open branch.] We only have one constant, ‘a’. So our universal decompositions have made all substitutions for constants in the branches. Thus even though we have not checked off our universally quantified formulas, we have completed the tree, and since all the branches are open, it is a completed open tree (Agler 296a).

In PL, the basic unit of representation is the proposition, which is assigned a truth value (true or false). In RL, the truth or falsity of a predicate well-formed formula (wff, pronounced ‘woof’) is relative to an interpretation in a model (i.e., relative to a specification of the domain of discourse and an interpretation function). Using the notion of an interpretation, semantic properties can be defined for RL propositions, sets of propositions, and arguments as follows:Tautology: A proposition ‘P’ is a tautology in RL if and only if ‘P’ is true on every interpretation.Contradiction: A proposition ‘P’ is a contradiction in RL if and only if ‘P’ is false on every interpretation.Contingency: A proposition ‘P’ is a contingency in RL if and only if ‘P’ is neither a contradiction nor a tautology.Equivalence Propositions: ‘P’ and ‘Q’ are equivalent in RL if and only if there is no interpretation where the valuation of ‘P’ is different from the valuation of ‘Q.’Consistency: A set of propositions ‘{A,B,C, ...,Z}’ is consistent in RL if and only if there is at least one interpretation such that all of the propositions in the set are true.Validity: An argument ‘P,Q,R, ...,Y⊢Z’ is valid in RL if and only if there is no interpretation such that all of the premises ‘A,’ ‘B,’ ‘C,’ ..., ‘Y’ are true and the conclusion ‘Z’ is false.(Agler 296)

(∀x)Px(∃x)Rx(Agler 296)

**Consistency**: A set of propositions ‘{

**A**,

**B**,

**C**, ...,

**Z**}’ is consistent in RL if and only if there is at least one interpretation such that all of the propositions in the set are true.”] This means we need to find at least one interpretation that makes both formulas true. [If we make the domain be all positive integers, and we furthermore make the predicate P mean “is greater than 0” and R “is even”, then we have found a domain and interpretation function that makes both propositions true. For, all positive integers are greater than zero, and there is at least one (in fact infinitely many) that are even.]

to show that ‘{(∀x)Px, (∃x)Rx}’ is consistent in RL involves showing that there is at least one interpretation in a model wherev(∀x)Px = T andv(∃x)Rx = T. Here is an example of such a model:D = positive integersP = {x|xis greater than 0}R = {x|xis even}(Agler 296)

1 | (∃x)Px | P |

2 | Pa | P |

1 | (∃x)Px | P |

2 | Pa | P |

3 | Pb | 1∃D |

**Completed open branch**: A branch is a completed open branch if and only if (1) all complex propositions that can be decomposed into atomic propositions or negated atomic propositions are decomposed; (2) for all universally quantified propositions ‘(∀x)

**P**’ occurring in the branch, there is a substitution instance ‘

**P**(a/

*x*)’ for each constant that occurs in that branch; and (3) the branch is not a closed branch.”] This branch is then completed open.

1 | (∃x)Px | P |

2 | Pa | P |

3 | Pb O | 1∃D |

we would stipulate a domain of discourse involving two objects, letting ‘a’ stand for an object and ‘b’ stand for an object, and assign the one-place predicate ‘P’ an extension.D: {John, Fred}Px:xis a person {John, Fred}a: Johnb: Fred(Agler 297)

*x*to either John or Fred. This means that Pa and Pb are true, because both ‘a’ and ‘b’ are people. That furthermore means that (∃x)Px is true, because there is at least one item in the domain which takes the predicate P.

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px | P |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

4 | Pa | | 3∃D |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

4 | Pa | | 3∃D |

5 | Pa | | 1∀D |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

4 | Pa | | 3∃D |

5 | Pa | O | 1∀D |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

4 | Pa | | 3∃D |

5 | Pa | O | | 1∀D |

6 | (∀x)¬Px | 3¬∃D |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

4 | Pa | | 3∃D |

5 | Pa | O | | 1∀D |

6 | (∀x)¬Px | 3¬∃D |

7 | ¬Pa | 6∀D |

8 | Pa | 1∀D |

1 | (∀x)Px | P |

2 | (∃x)Px∨¬(∃x)Px / \ | P |

3 | (∃x)Px ¬(∃x)Px | 2∨D |

4 | Pa | | 3∃D |

5 | Pa | O | | 1∀D |

6 | (∀x)¬Px | 3¬∃D |

7 | ¬Pa | 6∀D |

8 | Pa X | 1∀D |

D: {1}Px:xis a number(Agler 298)

1 | (∃x)(∃y)[(Ox∧Ey)∧Gxy] | P |

1 | (∃x)(∃y)[(Ox∧Ey)∧Gxy]✔ | P |

2 | (∃y)[(Oa∧Ey)∧Gay] | 1∃D |

1 | (∃x)(∃y)[(Ox∧Ey)∧Gxy]✔ | P |

2 | (∃y)[(Oa∧Ey)∧Gay]✔ | 1∃D |

3 | (Oa∧Eb)∧Gab | 2∃D |

1 | (∃x)(∃y)[(Ox∧Ey)∧Gxy]✔ | P |

2 | (∃y)[(Oa∧Ey)∧Gay]✔ | 1∃D |

3 | (Oa∧Eb)∧Gab✔ | 2∃D |

4 | Oa∧Eb✔ | 3∧D |

5 | Gab | 3∧D |

6 | Oa | 4∧D |

7 | Eb | 4∧D |

1 | (∃x)(∃y)[(Ox∧Ey)∧Gxy]✔ | P |

2 | (∃y)[(Oa∧Ey)∧Gay]✔ | 1∃D |

3 | (Oa∧Eb)∧Gab✔ | 2∃D |

4 | Oa∧Eb✔ | 3∧D |

5 | Gab | 3∧D |

6 | Oa | 4∧D |

7 | Eb O | 4∧D |

there is an interpretation for which the propositions ‘Gab,’ ‘Oa,’ and ‘Eb’ are true, and thus ‘(∃x)(∃y) [(Ox∧Ey)∧Gxy]’ is true. Again, a model can be constructed to reflect this fact:D: {1, 2, 3}Ox:xis an odd numberEx:xis an even numberGxy:xis greater thanya: 3b: 2‘Gab’ is true since it is true that 3 is greater than 2. ‘Oa’ is true since three is an odd number. Lastly, ‘Eb’ is true since 2 is an even number. Thus, the predicate wff ‘(∃x)(∃y)[(Ox∧Ey)∧Gxy],’ which says that there exists an odd number greater than | some existent even number, is also true. In short, the truth tree, along with the model, demonstrates that the set ‘{(∃x)(∃y)[(Ox∧Ey)∧Gxy]}’ is not a contradiction in RL.(Agler 298-299)

Consistency: A set of propositions ‘{P,Q,R, ...,Z}’ is shown by the truth-tree method to be consistent if and only if a tree of the stack ‘P,’ ‘Q,’ ‘R,’ . . ., ‘Z’ is an open tree; that is, there is at least one completed open branch.Inconsistency: A set of propositions ‘{P,Q,R, ...,Z}’ is shown by the truth-tree method to be inconsistent if and only if a tree of the stack of ‘P,’ ‘Q,’ ‘R,’ . . ., ‘Z’ is a closed tree; that is, all branches close.

(Agler 299)

(∀x)(Px→Rx), ¬(∀x)(¬Rx→¬Px)(Agler 299)

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px) | P |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px) | 2¬∀D |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px)✔ | 2¬∀D |

4 | ¬(¬Ra→¬Pa) | 3∃D |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px)✔ | 2¬∀D |

4 | ¬(¬Ra→¬Pa) | 3∃D |

5 | Pa→Ra | 1∀D |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px)✔ | 2¬∀D |

4 | ¬(¬Ra→¬Pa) | 3∃D |

5 | Pa→Ra✔ | 1∀D |

6 | ¬Ra | 4¬→D |

7 | ¬¬Pa | 4¬→D |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px)✔ | 2¬∀D |

4 | ¬(¬Ra→¬Pa) | 3∃D |

5 | Pa→Ra✔ | 1∀D |

6 | ¬Ra | 4¬→D |

7 | ¬¬Pa✔ | 4¬→D |

8 | Pa | 7¬¬D |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px)✔ | 2¬∀D |

4 | ¬(¬Ra→¬Pa) | 3∃D |

5 | Pa→Ra✔ | 1∀D |

6 | ¬Ra | 4¬→D |

7 | ¬¬Pa✔ | 4¬→D |

8 | Pa / \ | 7¬¬D |

9 | ¬Pa Ra X X | 5→D |

1 | (∀x)(Px→Rx) | P |

2 | ¬(∀x)(¬Rx→¬Px)✔ | P |

3 | (∃x)¬(¬Rx→¬Px)✔ | 2¬∀D |

4 | ¬(¬Ra→¬Pa) | 3∃D |

5 | Pa→Ra✔ | 1∀D |

6 | ¬Ra | 4¬→D |

7 | ¬¬Pa✔ | 4¬→D |

8 | Pa / \ | 7¬¬D |

9 | ¬Pa Ra | 5→D |

1 | (∀x)(Px)→(∀y)(Ry) | P |

2 | ¬(∀y)Ry | P |

3 | (∃x)¬Px | P |

1 | (∀x)(Px)→(∀y)(Ry) | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px | P |

4 | (∃y)¬Ry | 2¬∀D |

1 | (∀x)(Px)→(∀y)(Ry) | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb | 3∃D |

1 | (∀x)(Px)→(∀y)(Ry)✔ | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb / \ | 3∃D |

7 | ¬(∀x)Px (∀y)Ry | 1→D |

1 | (∀x)(Px)→(∀y)(Ry)✔ | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb / \ | 3∃D |

7 | ¬(∀x)Px✔ (∀y)Ry | 1→D |

8 | (∃x)¬Px | | 7¬∀D |

1 | (∀x)(Px)→(∀y)(Ry)✔ | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb / \ | 3∃D |

7 | ¬(∀x)Px✔ (∀y)Ry | 1→D |

8 | (∃x)¬Px✔ | | 7¬∀D |

9 | ¬Pc | | 8∃D |

1 | (∀x)(Px)→(∀y)(Ry)✔ | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb / \ | 3∃D |

7 | ¬(∀x)Px✔ (∀y)Ry | 1→D |

8 | (∃x)¬Px✔ | | 7¬∀D |

9 | ¬Pc | O | | 8∃D |

1 | (∀x)(Px)→(∀y)(Ry)✔ | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb / \ | 3∃D |

7 | ¬(∀x)Px✔ (∀y)Ry | 1→D |

8 | (∃x)¬Px✔ | | 7¬∀D |

9 | ¬Pc | O | | 8∃D |

10 | Ra | 7∀D |

11 | Rb | 7∀D |

1 | (∀x)(Px)→(∀y)(Ry)✔ | P |

2 | ¬(∀y)Ry✔ | P |

3 | (∃x)¬Px✔ | P |

4 | (∃y)¬Ry✔ | 2¬∀D |

5 | ¬Ra | 4∃D |

6 | ¬Pb / \ | 3∃D |

7 | ¬(∀x)Px✔ (∀y)Ry | 1→D |

8 | (∃x)¬Px✔ | | 7¬∀D |

9 | ¬Pc | O | | 8∃D |

10 | Ra | 7∀D |

11 | Rb X | 7∀D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy) | P |

2 | ¬(∀y)(∀x)(Rxy) | P |

3 | (Rab∧Rba)∧Pab | P |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy) | P |

2 | ¬(∀y)(∀x)(Rxy) | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy) | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy) | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy) | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy) | 2¬∀D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy) | 2¬∀D |

12 | ¬(∃y)(Pcy) | 10∃D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy) | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy) | 12¬∃D |

*x*place. And we have three constants to substitute in for

*y*, namely, ‘a’, ‘b’, and ‘c’.]

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy) | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy)✔ | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

17 | ¬(∀x)(Rxe) | 11∃D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy)✔ | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

17 | ¬(∀x)(Rxe)✔ | 11∃D |

18 | (∃x)¬(Rxe) | 17¬∀D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy) | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy)✔ | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

17 | ¬(∀x)(Rxe)✔ | 11∃D |

18 | (∃x)¬(Rxe)✔ | 17¬∀D |

19 | ¬Rfe | 18∃D |

*y*.]

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy)✔ | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy)✔ | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

17 | ¬(∀x)(Rxe)✔ | 11∃D |

18 | (∃x)¬(Rxe)✔ | 17¬∀D |

19 | ¬Rfe | 18∃D |

20 | ¬(∃x)(Rxa) | 9∀D |

21 | ¬(∃x)(Rxb) | 9∀D |

22 | ¬(∃x)(Rxe) | 9∀D |

23 | ¬(∃x)(Rxf) | 9∀D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy)✔ | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy)✔ | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

17 | ¬(∀x)(Rxe)✔ | 11∃D |

18 | (∃x)¬(Rxe)✔ | 17¬∀D |

19 | ¬Rfe | 18∃D |

20 | ¬(∃x)(Rxa)✔ | 9∀D |

21 | ¬(∃x)(Rxb)✔ | 9∀D |

22 | ¬(∃x)(Rxe)✔ | 9∀D |

23 | ¬(∃x)(Rxf)✔ | 9∀D |

24 | (∀x)¬(Rxa) | 20¬∃D |

25 | (∀x)¬(Rxb) | 21¬∃D |

26 | (∀x)¬(Rxe) | 22¬∃D |

27 | (∀x)¬(Rxf) | 23¬∃D |

1 | ¬(∀x)(∃y)(Pxy)∧(∀y)¬(∃x)(Rxy)✔ | P |

2 | ¬(∀y)(∀x)(Rxy)✔ | P |

3 | (Rab∧Rba)∧Pab ✔ | P |

4 | Rab∧Rba✔ | 3∧D |

5 | Pab | 3∧D |

6 | Rab | 4∧D |

7 | Rba | 4∧D |

8 | ¬(∀x)(∃y)(Pxy)✔ | 1∧D |

9 | (∀y)¬(∃x)(Rxy)✔ | 1∧D |

10 | (∃x)¬(∃y)(Pxy)✔ | 8¬∀D |

11 | (∃y)¬(∀x)(Rxy)✔ | 2¬∀D |

12 | ¬(∃y)(Pcy)✔ | 10∃D |

13 | (∀y)¬(Pcy)✔ | 12¬∃D |

14 | ¬Pca | 13∀D |

15 | ¬Pcb | 13∀D |

16 | ¬Pcc | 13∀D |

17 | ¬(∀x)(Rxe)✔ | 11∃D |

18 | (∃x)¬(Rxe)✔ | 17¬∀D |

19 | ¬Rfe | 18∃D |

20 | ¬(∃x)(Rxa)✔ | 9∀D |

21 | ¬(∃x)(Rxb)✔ | 9∀D |

22 | ¬(∃x)(Rxe)✔ | 9∀D |

23 | ¬(∃x)(Rxf)✔ | 9∀D |

24 | (∀x)¬(Rxa) | 20¬∃D |

25 | (∀x)¬(Rxb) | 21¬∃D |

26 | (∀x)¬(Rxe) | 22¬∃D |

27 | (∀x)¬(Rxf) | 23¬∃D |

28 | ¬Rab | 25∀D |

Tautology: A proposition ‘P’ is shown by the truth-tree method to be a tautology if and only if the tree ‘¬P’ determines a closed tree; that is, all branches close.Contradiction: A proposition ‘P’ is shown by the truth-tree method to be a contradiction if and only if the tree ‘P’ determines a closed tree; that is, all branches close.Contingency: A proposition ‘P’ is shown by the truth-tree method to be a contingency if and only if ‘P’ is neither a tautology nor a contradiction; that is, the tree of ‘P’ does not determine a closed tree, and the tree of ‘¬P’ does not determine a closed tree.(Agler 305)

“To summarize, the truth-tree method can be used to determine whether a proposition ‘

**P**’ is a tautology, contradiction, or contingency. In testing ‘

**P**’ to see if it is a tautology, begin the tree with ‘¬

**P**.’ If the tree closes, you know that it is a tautology. If the tree is open, then ‘

**P**’ is either a contradiction or a contingency. Similarly, in testing ‘

**P**’ to see if it is a contradiction, begin the tree with ‘

**P**.’ If the tree closes, you know that it is a contradiction. If the tree is open, then ‘

**P**’ is either a tautology or a contingency. Lastly, if the truth-tree test shows that ‘

**P**’ is neither a contradiction nor a tautology, then ‘

**P**’ is a contingency” (Agler 147). Perhaps another way would be first to test if it is a contingency by seeing if it makes a closed tree. If the tree is closed, then it is a contradiction. If it is not, then we do another test. We negate the proposition and see if it makes a closed tree. If it does, then it is a tautology. If it does not, then it is a contingency.] We will now look at a proposition and see if it is a contradiction, which we do by seeing if all the branches close.

1 | (∃x)¬(∀y)[Px→(Qx∨¬Ry)] | p |

1 | (∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | p |

2 | ¬(∀y)[Pa→(Qa∨¬Ry)] | 1∃D |

1 | (∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | p |

2 | ¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 1∃D |

3 | (∃y)¬[Pa→(Qa∨¬Ry)] | 2¬∀D |

1 | (∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | p |

2 | ¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 1∃D |

3 | (∃y)¬[Pa→(Qa∨¬Ry)]✔ | 2¬∀D |

4 | ¬[Pa→(Qa∨¬Rb)] | 3∃D |

1 | (∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | p |

2 | ¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 1∃D |

3 | (∃y)¬[Pa→(Qa∨¬Ry)]✔ | 2¬∀D |

4 | ¬[Pa→(Qa∨¬Rb)]✔ | 3∃D |

5 | Pa | 4¬→D |

6 | ¬(Qa∨¬Rb) | 4¬→D |

1 | (∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | p |

2 | ¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 1∃D |

3 | (∃y)¬[Pa→(Qa∨¬Ry)]✔ | 2¬∀D |

4 | ¬[Pa→(Qa∨¬Rb)]✔ | 3∃D |

5 | Pa | 4¬→D |

6 | ¬(Qa∨¬Rb)✔ | 4¬→D |

7 | ¬Qa | 6¬∨D |

8 | ¬¬Rb | 6¬∨D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)] | P |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)] | 1¬∃D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)] | 2∀D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 2∀D |

4 | (∀y)[Pa→(Qa∨¬Ry)] | 3¬¬D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 2∀D |

4 | (∀y)[Pa→(Qa∨¬Ry)] | 3¬¬D |

5 | Pa→(Qa∨¬Ra) | 4∀D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 2∀D |

4 | (∀y)[Pa→(Qa∨¬Ry)] | 3¬¬D |

5 | Pa→(Qa∨¬Ra) / \ | 4∀D |

6 | ¬Pa Qa∨¬Ra | 5→D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 2∀D |

4 | (∀y)[Pa→(Qa∨¬Ry)] | 3¬¬D |

5 | Pa→(Qa∨¬Ra) / \ | 4∀D |

6 | ¬Pa Qa∨¬Ra O | 5→D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 2∀D |

4 | (∀y)[Pa→(Qa∨¬Ry)] | 3¬¬D |

5 | Pa→(Qa∨¬Ra) / \ | 4∀D |

6 | ¬Pa Qa∨¬Ra✔ O / \ | 5→D |

7 | Qa ¬Ra | 6∨D |

1 | ¬(∃x)¬(∀y)[Px→(Qx∨¬Ry)]✔ | P |

2 | (∀x)¬¬(∀y)[Px→(Qx∨¬Ry)]✔ | 1¬∃D |

3 | ¬¬(∀y)[Pa→(Qa∨¬Ry)]✔ | 2∀D |

4 | (∀y)[Pa→(Qa∨¬Ry)] | 3¬¬D |

5 | Pa→(Qa∨¬Ra) / \ | 4∀D |

6 | ¬Pa Qa∨¬Ra✔ O / \ | 5→D |

7 | Qa ¬Ra O O | 6∨D |

Agler will perform these tests for the following proposition.

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx) | P |

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx)✔ | P |

2 | (∀x)(Px→Qx) | 1∧D |

3 | (∃x)(Px∧¬Qx) | 1∧D |

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx)✔ | P |

2 | (∀x)(Px→Qx) | 1∧D |

3 | (∃x)(Px∧¬Qx)✔ | 1∧D |

4 | Pa∧¬Qa | 3∃D |

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx)✔ | P |

2 | (∀x)(Px→Qx) | 1∧D |

3 | (∃x)(Px∧¬Qx)✔ | 1∧D |

4 | Pa∧¬Qa✔ | 3∃D |

5 | Pa | 4∧D |

6 | ¬Qa | 4∧D |

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx)✔ | P |

2 | (∀x)(Px→Qx) | 1∧D |

3 | (∃x)(Px∧¬Qx)✔ | 1∧D |

4 | Pa∧¬Qa✔ | 3∃D |

5 | Pa | 4∧D |

6 | ¬Qa | 4∧D |

7 | Pa→Qa | 2∀D |

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx)✔ | P |

2 | (∀x)(Px→Qx) | 1∧D |

3 | (∃x)(Px∧¬Qx)✔ | 1∧D |

4 | Pa∧¬Qa✔ | 3∃D |

5 | Pa | 4∧D |

6 | ¬Qa | 4∧D |

7 | Pa→Qa✔ / \ | 2∀D |

8 | ¬Pa Qa | 7→D |

1 | (∀x)(Px→Qx)∧(∃x)(Px∧¬Qx)✔ | P |

2 | (∀x)(Px→Qx) | 1∧D |

3 | (∃x)(Px∧¬Qx)✔ | 1∧D |

4 | Pa∧¬Qa✔ | 3∃D |

5 | Pa | 4∧D |

6 | ¬Qa | 4∧D |

7 | Pa→Qa✔ / \ | 2∀D |

8 | ¬Pa Qa X X | 7→D |

(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)(Agler 307)

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)] | P |

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)]✔ / \ | P |

2 | ¬(∀x)(Px→Px) ¬(∀y)(Qy∨¬Qy) | 1¬∧D |

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)]✔ / \ | P |

2 | ¬(∀x)(Px→Px)✔ ¬(∀y)(Qy∨¬Qy✔ | 1¬∧D |

3 | (∃x)¬(Px→Px) (∃y)¬(Qy∨¬Qy) | 2¬∀D |

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)]✔ / \ | P |

2 | ¬(∀x)(Px→Px)✔ ¬(∀y)(Qy∨¬Qy✔ | 1¬∧D |

3 | (∃x)¬(Px→Px)✔ (∃y)¬(Qy∨¬Qy)✔ | 2¬∀D |

4 | ¬(Pa→Pa) ¬(Qa∨¬Qa) | 3∃D |

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)]✔ / \ | P |

2 | ¬(∀x)(Px→Px)✔ ¬(∀y)(Qy∨¬Qy✔ | 1¬∧D |

3 | (∃x)¬(Px→Px)✔ (∃y)¬(Qy∨¬Qy)✔ | 2¬∀D |

4 | ¬(Pa→Pa) ¬(Qa∨¬Qa) | 3∃D |

5 | Pa | | 4¬→D |

6 | ¬Pa | | 4¬→D |

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)]✔ / \ | P |

2 | ¬(∀x)(Px→Px)✔ ¬(∀y)(Qy∨¬Qy✔ | 1¬∧D |

3 | (∃x)¬(Px→Px)✔ (∃y)¬(Qy∨¬Qy)✔ | 2¬∀D |

4 | ¬(Pa→Pa) ¬(Qa∨¬Qa) | 3∃D |

5 | Pa | | 4¬→D |

6 | ¬Pa | X | | 4¬→D |

7 | ¬Qa | 4¬∨D |

8 | ¬¬Qa | 4¬∨D |

9 | Qa | 8¬¬D |

1 | ¬[(∀x)(Px→Px)∧(∀y)(Qy∨¬Qy)]✔ / \ | P |

2 | ¬(∀x)(Px→Px)✔ ¬(∀y)(Qy∨¬Qy✔ | 1¬∧D |

3 | (∃x)¬(Px→Px)✔ (∃y)¬(Qy∨¬Qy)✔ | 2¬∀D |

4 | ¬(Pa→Pa) ¬(Qa∨¬Qa) | 3∃D |

5 | Pa | | 4¬→D |

6 | ¬Pa | X | | 4¬→D |

7 | ¬Qa | 4¬∨D |

8 | ¬¬Qa | 4¬∨D |

9 | Qa X | 8¬¬D |

7.3.4 Logical Equivalence

Equivalence: A pair of propositions ‘P’ and ‘Q’ is shown by the truth-tree method to be equivalent if and only if the tree of the stack of ‘¬(P↔Q)’ determines a closed tree; that is, all branches for ‘¬(P↔Q)’ close.(Agler 310)

(∀x)Px¬(∃x)Px(Agler 310)

1 | ¬[(∀x)Px↔¬(∃x)Px] | P |

1 | ¬[(∀x)Px↔¬(∃x)Px]✔ / \ | P |

2 | (∀x)Px ¬(∀x)Px | 1¬↔D |

3 | ¬¬(∃x)Px ¬(∃x)Px | 1¬↔D |

1 | ¬[(∀x)Px↔¬(∃x)Px]✔ / \ | P |

2 | (∀x)Px ¬(∀x)Px | 1¬↔D |

3 | ¬¬(∃x)Px✔ ¬(∃x)Px | 1¬↔D |

4 | (∃x)Px | 3¬¬D |

1 | ¬[(∀x)Px↔¬(∃x)Px]✔ / \ | P |

2 | (∀x)Px ¬(∀x)Px | 1¬↔D |

3 | ¬¬(∃x)Px✔ ¬(∃x)Px | 1¬↔D |

4 | (∃x)Px✔ | 3¬¬D |

5 | Pa | 4∃D |

1 | ¬[(∀x)Px↔¬(∃x)Px]✔ / \ | P |

2 | (∀x)Px ¬(∀x)Px | 1¬↔D |

3 | ¬¬(∃x)Px✔ ¬(∃x)Px | 1¬↔D |

4 | (∃x)Px✔ | 3¬¬D |

5 | Pa | 4∃D |

6 | Pa | 2∀D |

1 | ¬[(∀x)Px↔¬(∃x)Px]✔ / \ | P |

2 | (∀x)Px ¬(∀x)Px | 1¬↔D |

3 | ¬¬(∃x)Px✔ ¬(∃x)Px | 1¬↔D |

4 | (∃x)Px✔ | 3¬¬D |

5 | Pa | 4∃D |

6 | Pa O | 2∀D |

(∀x)¬(Px∨Gx)(∀y)(¬Py∧¬Gy)(Agler 310)

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]} | P |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx) (∀y)(¬Px∧¬Gx) | 1¬↔D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx) | | 3¬∀D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga) | | 4∃D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa ¬¬Ga | | 5¬∧D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga) ¬(Pa∨Ga) | | 2∀D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga) ¬(Pa∨Ga) | | 2∀D |

9 | ¬Pa ¬Pa | | 8¬∨D |

10 | X ¬Ga |X | | 8¬∨D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga)✔ ¬(Pa∨Ga)✔ | | 2∀D |

9 | ¬Pa ¬Pa | | 8¬∨D |

10 | X ¬Ga |X | | 8¬∨D |

11 | (∃x)¬¬(Px∨Gx) | 2¬∀D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga)✔ ¬(Pa∨Ga)✔ | | 2∀D |

9 | ¬Pa ¬Pa | | 8¬∨D |

10 | X ¬Ga |X | | 8¬∨D |

11 | (∃x)¬¬(Px∨Gx)✔ | 2¬∀D |

12 | Pa∨Ga | 11∃D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga)✔ ¬(Pa∨Ga)✔ | | 2∀D |

9 | ¬Pa ¬Pa | | 8¬∨D |

10 | X ¬Ga |X | | 8¬∨D |

11 | (∃x)¬¬(Px∨Gx)✔ | 2¬∀D |

12 | Pa∨Ga✔ / \ | 11∃D |

13 | Pa Ga | 12∨D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga)✔ ¬(Pa∨Ga)✔ | | 2∀D |

9 | ¬Pa ¬Pa | | 8¬∨D |

10 | X ¬Ga |X | | 8¬∨D |

11 | (∃x)¬¬(Px∨Gx)✔ | 2¬∀D |

12 | Pa∨Ga✔ / \ | 11∃D |

13 | Pa Ga | 12∨D |

14 | ¬Pa∧¬Ga ¬Pa∧¬Ga | 3∀D |

1 | ¬{[(∀x)¬(Px∨Gx)]↔[(∀y)(¬Py∧¬Gy)]}✔ / \ | P |

2 | (∀x)¬(Px∨Gx) ¬(∀x)¬(Px∨Gx) | 1¬↔D |

3 | ¬(∀y)(¬Px∧¬Gx)✔ (∀y)(¬Px∧¬Gx) | 1¬↔D |

4 | (∃y)¬(¬Px∧¬Gx)✔ | | 3¬∀D |

5 | ¬(¬Pa∧¬Ga)✔ | / \ | | 4∃D |

6 | ¬¬Pa✔ ¬¬Ga✔ | | 5¬∧D |

7 | Pa Ga | | 6¬¬D |

8 | ¬(Pa∨Ga)✔ ¬(Pa∨Ga)✔ | | 2∀D |

9 | ¬Pa ¬Pa | | 8¬∨D |

10 | X ¬Ga |X | | 8¬∨D |

11 | (∃x)¬¬(Px∨Gx)✔ | 2¬∀D |

12 | Pa∨Ga✔ / \ | 11∃D |

13 | Pa Ga | 12∨D |

14 | ¬Pa∧¬Ga ¬Pa∧¬Ga | 3∀D |

15 | ¬Pa ¬Ga X X | 14∧D |

Validity: An argument ‘P,Q,R, ...,Y⊢Z’ is shown by the truth-tree method to be valid in RL if and only if the stack ‘P,’ ‘Q,’ ‘R,’ ..., ‘Y,’ ‘¬Z’ determines a closed tree.Invalidity: An argument ‘P,Q,R, ...,Y⊢Z’ is shown by the truth-tree method to be invalid in RL if and only if the stack ‘P,’ ‘Q,’ ‘R,’ ..., ‘Y,’ ‘¬Z’ has at least one completed open branch. (Agler 314)

Agler first shows how we do this with a simple argument, namely:

(∀x)Px, Pa ⊢ Pa

(Agler 314)

1 | (∀x)Px | P |

2 | ¬Pa | P |

1 | (∀x)Px | P |

2 | ¬Pa | P |

3 | Pa X | 1∀D |

(∀x)(Px→Qx), (∃y)(Py) ⊢ (∃x)(Px→Qx)(Agler 314)

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py) | P |

3 | ¬(∃x)(Px→Qx) | P |

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py)✔ | P |

3 | ¬(∃x)(Px→Qx) | P |

4 | Pa | 2∃D |

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py)✔ | P |

3 | ¬(∃x)(Px→Qx)✔ | P |

4 | Pa | 2∃D |

5 | (∀x)¬(Px→Qx) | 3¬∃D |

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py)✔ | P |

3 | ¬(∃x)(Px→Qx)✔ | P |

4 | Pa | 2∃D |

5 | (∀x)¬(Px→Qx) | 3¬∃D |

6 | Pa→Qa | 1∀D |

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py)✔ | P |

3 | ¬(∃x)(Px→Qx)✔ | P |

4 | Pa | 2∃D |

5 | (∀x)¬(Px→Qx) | 3¬∃D |

6 | Pa→Qa | 1∀D |

7 | ¬(Pa→Qa) | 5∀D |

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py)✔ | P |

3 | ¬(∃x)(Px→Qx)✔ | P |

4 | Pa | 2∃D |

5 | (∀x)¬(Px→Qx) | 3¬∃D |

6 | Pa→Qa | 1∀D |

7 | ¬(Pa→Qa)✔ | 5∀D |

8 | Pa | 7¬→D |

9 | ¬Qa | 7¬→D |

1 | (∀x)(Px→Qx) | P |

2 | (∃y)(Py)✔ | P |

3 | ¬(∃x)(Px→Qx)✔ | P |

4 | Pa | 2∃D |

5 | (∀x)¬(Px→Qx) | 3¬∃D |

6 | Pa→Qa | 1∀D |

7 | ¬(Pa→Qa)✔ | 5∀D |

8 | Pa | 7¬→D |

9 | ¬Qa / \ | 7¬→D |

10 | ¬Pa Qa X X | 6→D |

Pa∧Qb, (∀x)(∀y)[Px→(Qy→Rx)] ⊢ Ra(Agler 315)

{Pa∧Qb, (∀x)(∀y)[Px→(Qy→Rx)], ¬Ra}(Agler 315)

1 | Pa∧Qb | P |

2 | (∀x)(∀y)[Px→(Qy→Rx)] | P |

3 | ¬Ra | P |

1 | Pa∧Qb✔ | P |

2 | (∀x)(∀y)[Px→(Qy→Rx)] | P |

3 | ¬Ra | P |

4 | Pa | 1∧D |

5 | Qb | 1∧D |

1 | Pa∧Qb✔ | P |

2 | (∀x)(∀y)[Px→(Qy→Rx)] | P |

3 | ¬Ra | P |

4 | Pa | 1∧D |

5 | Qb | 1∧D |

6 | (∀y)[Pa→(Qy→Ra)] | 2∀D |

7 | (∀y)[Pb→(Qy→Rb)] | 2∀D |

1 | Pa∧Qb✔ | P |

2 | (∀x)(∀y)[Px→(Qy→Rx)] | P |

3 | ¬Ra | P |

4 | Pa | 1∧D |

5 | Qb | 1∧D |

6 | (∀y)[Pa→(Qy→Ra)] | 2∀D |

7 | (∀y)[Pb→(Qy→Rb)] | 2∀D |

8 | Pa→(Qa→Ra) | 6∀D |

9 | Pa→(Qb→Ra) | 6∀D |

10 | Pb→(Qa→Rb) | 7∀D |

11 | Pb→(Qb→Rb) | 7∀D |

1 | Pa∧Qb✔ | P |

2 | (∀x)(∀y)[Px→(Qy→Rx)] | P |

3 | ¬Ra | P |

4 | Pa | 1∧D |

5 | Qb | 1∧D |

6 | (∀y)[Pa→(Qy→Ra)] | 2∀D |

7 | (∀y)[Pb→(Qy→Rb)] | 2∀D |

8 | Pa→(Qa→Ra) | 6∀D |

9 | Pa→(Qb→Ra)✔ | 6∀D |

10 | Pb→(Qa→Rb) | 7∀D |

11 | Pb→(Qb→Rb) / \ | 7∀D |

12 | ¬Pa Qb→Ra✔ X / \ | 9→D |

13 | ¬Qb Ra X X | 12→D |

Pa∧Qb, (∀x)(∀y)[Px→(Qy→Rx)], ¬Rais inconsistent. That furthermore means that the argument

(Agler 316)

Pa∧Qb, (∀x)(∀y)[Px→(Qy→Rx)] ⊢ Ra

*Symbolic Logic: Syntax, Semantics, and Proof*. New York: Rowman & Littlefield, 2013.

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