I will just post these questions now, and post my feeble attempts later.
2. Build a truth table to show that (A => B) <=> (not B or A) is true for all truth values of A and B.
3. Build a truth table to show that (A does not imply B) is equivalent to (A and not B) is a tautology.
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ReplyDelete(resubmitted because blogger ate my tabs...)
ReplyDelete2.
Φ | Ψ | ¬Φ | Φ ⇒ Ψ | ¬Φ ⋁ Ψ
T | T | F | T | T
T | F | F | F | F
F | T | T | T | T
F | F | T | T | T
3.
Φ | Ψ ¬Ψ | Φ ⇏ Ψ | Φ ⋀ ¬Ψ
T | T | F | F | F
T | F | T | T | T
F | T | F | F | F
F | F | T | F | F
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DeleteHaving second thoughts about tautology proof.
DeleteΦ|ψ|~ψ|(Φ => ψ)|(Φ =/> ψ)|(Φ ^ ~ψ)|(Φ =/> ψ) <=> (Φ^~ψ)
DeleteT|T|F | T | F | F | T
T|F|T | F | T | T | T
F|T|F | T | F | F | T
F|F|T | T | F | F | T
I believe the last column completes the proof for tautology.
I got the same answers
ReplyDeleteI carried the truth table out so that each of the implications was included. That gave two additional columns, from which one determines the truth of the proposition.
ReplyDeleteFWIW, Wikipedia calls Φ ⇏ Ψ a "material implication." The truth table is given as:
P Q P⇏Q
T T F
T F T
F T F
F F F
There also is a Venn diagram to show what the result is. See it here:
http://en.wikipedia.org/wiki/Material_nonimplication#Computer_Science