by Corry Shores
[Search Blog Here. Index-tags are found on the bottom of the left column.]
[Logic & Semantics, Entry Directory]
[David Agler, entry directory]
[Agler’s Symbolic Logic, entry directory]
[The following is summary. Boldface (except for metavariables) and bracketed commentary are my own. Please forgive my typos, as proofreading is incomplete. I highly recommend Agler’s excellent book. It is one of the best introductions to logic I have come across.]
Summary of
David W. Agler
Symbolic Logic: Syntax, Semantics, and Proof
Ch.4: Truth Trees
4.3 The Remaining Decomposition Rules
Brief summary:
With all the truth-tree decomposition rules covered, we may place them together in one table. 
Summary
4.3 The Remaining Decomposition Rules
Previously we formulated two of the decomposition rules for truth-trees, namely, the conjunction decomposition rule (∧D) and the disjunction decomposition rule (∨D). We now examine the remaining seven rules (120).
4.3.1 Conditional Decomposition (→D)
Recall (from section 2.4) the truth-table from the material conditional.
[Recall from section 4.1.2 that we use a branching rule when the proposition is false under just one truth-value assignment.] Since conditionals are false under only one value-assignment, namely, if v(P) = T and v(Q) = F, they will branch when we decompose them. But we need to know if we negate any of the terms at the end of each branch. To determine this, we need to know which forms of the constituent terms need to be true for the conditional to be true. [Look at the lines in the table where the conditional is true. We see that in both cases where P is false, the conditional is true. Since this exhausts all possibilities for P, that means, so long as P is false, the conditional will be true too. What pattern do we see for Q? In one case when it is false, the conditional is true, and in another case when it is false, the conditional is false. However, in both cases when Q is true, the conditional is true. Thus, a conditional is true so long as either P is false or Q is true. But at the end of each branch we can only put terms that would be true in order for the conditional to be true. So consider if we made the chart that also shows the values for ¬P .
Here we see that it is true whenever Q is true or whenever ¬P is true.] Given that
v(P→Q) = T if and only if v(¬P) = T or v(Q) = T
(Agler 121)
we place ¬P at the end of one branch and Q at the end of the other.
Agler will illustrate by decomposing the following formula:
P→(W→Z)
(Agler 121)
We first give it its own line, and place P for proposition.
We next decompose it using the branching rule for conditionals. We justify the decomposition with 1→D to indicate it came from line 1, and and we place a checkmark next to that formula above.
We finally decompose the right branch, following the same procedure.
4.3.2 Biconditional Decomposition (↔D)
Recall the table for the biconditional.
[And recall again from section 4.1.2 that we use a stacking and branching rule when the proposition is true under two truth-value assignments and false under two truth-value assignments.] To decompose the biconditional, we will use a combination of stacking and branching.
Agler illustrates with the tree for:
P↔(W∧Z)
(Agler 121)
We begin:
We then decompose the biconditional:
And finally we decompose the conjunction. At the moment we do not decompose the negated conjunction, since we still need the rule for it.
[Note, in my version of the text, there is a checkmark next to ¬P. I left it out of the diagram, because I did not understand why it was there.]
4.3.3 Negated Conjunction Decomposition (¬∧D)
[Consider the truth table for the negated conjunction.
And recall from section 4.1.2 that we use a branching rule when the proposition is false under just one truth-value assignment.] A negated conjunction is false whenever both conjuncts are true. [I might have this wrong. Agler writes: “A negated conjunction is only false in one case: when both of the conjuncts are false” (112). That might mean the negated conjunction is false when the conjunction of both terms is true.] He continues, “This means that a negated conjunction is true if either of the negated conjuncts is true”
and thus ¬∧D is a branching rule (122).
Agler illustrates by returning to our example for P↔(W∧Z). Again, we first begin with this proposition.
[Note, in my version of the text, it is shown as a conditional.] Next, we decompose the biconditional.
We then decompose the conjunction.
And finally, we decompose the negated conjunction.
[Note, in my version of the text, there is a checkmark next to ¬P, but I left it out, as I did not understand why it was there. I also added a ‘6’ for the sixth line.]
4.3.4 Negated Disjunction Decomposition (¬∨D)
[Consider the truth table for negated disjunction.
As we can see, it is true under just one assignment, namely, when both disjuncts are false. Recall from section 4.1.2 that we use a stacking rule when the proposition is true under just one truth-value assignment.] Agler writes:
A negated disjunction is only true in one case: when both of its constituents are false. That is, a negated disjunction is true if and only if both of the negated disjuncts are true. Therefore, it is a stacking rule.
(123)
[Note that the negated disjunction is true when both the negation of one conjunct and the negation of the other conjunct are jointly true.]
Agler illustrates with the example
¬[Z∨(B∨R)]
(Agler 123)
We start by placing it as a proposition.
We then decompose the negated disjunction.
And we finally decompose the new negated disjunction.
4.3.5 Negated Conditional Decomposition (¬→D)
[Consider the truth table for the negated conditional.
As we can see, it is true under just one assignment, and it is thus a stacking rule. When it is true, the terms that would also be true are P and ¬Q.
] Agler writes:
A negated conditional is true in one case: if and only if the both the antecedent is true and the consequent is false. Therefore, it is a stacking rule.
(Agler 123)
He illustrates by decomposing
¬[L→(B→R)]
(Agler 123)
First we place it as the proposition.
Next we decompose the negated conditional.
[Note, in my version of the text, the rule for line 3 reads “1¬∨D”.]
Then we decompose the negated conditional in line 3.
[Note, in my version of the text, the rule for lines 4 and 5 read “3¬∨D”.]
4.3.6 Negated Biconditional Decomposition (¬↔D)
[Consider the truth-table for the negated biconditional.
As we can see, it is true under two assignments and false under two assignments. It therefore takes a stacking and branching rule. And as we can see, when it is true, either P and ¬Q are each true, or ¬P and Q are each true.
] Agler writes: “A negated biconditional is true in two cases and false in two cases. Therefore, it both stacks and branches” (123).
He illustrates with
¬[P↔(W∨Z)]
(Agler 124)
First we place the proposition.
Then we decompose the negated biconditional.
Now let us decompose the negated disjunction on the left branch.
And finally, we decompose the disjunction on the right branch.
[Note, in my version of the text, there is a checkmark next to the ¬P in line 2.]
4.3.7 Double Negation Decomposition (¬¬D)
[Consider the truth-table for double negation.
As we can see, it is true under just one assignment, which means it takes a stacking rule. And when it is true, the term that would be true is P itself.] Agler writes: “A proposition that is doubly negated is true in one case, when the proposition in unnegated form is true. Therefore, it stacks” (124).
He illustrates with the following formula:
¬(¬P∨¬Z)
(Agler 124)
First we place it as a proposition.
We then decompose the negated disjunction.
We next decompose the double negation of line 2.
And finally we decompose the double negation of line 3.
4.3.8 Truth-Tree Rules
Agler displays a chart for all the truth-tree decomposition rules (125).
Agler, David. Symbolic Logic: Syntax, Semantics, and Proof. New York: Rowman & Littlefield, 2013.
.
No comments:
Post a Comment