However, suppose formula S (but I mistakenly wrote R) is a label for (A˄(B˅~B)). Any valuation that makes A false (when evaluating per row), will make the formula S false, so that S is not a tautology (as a formula).

R→S. S is a label for (A˄(B˅~B)), which is not a tautology |1100|. Then, if R is a tautology, then R→S is not a tautology |1100| (as a formula).

(A ˄ (B ˅ ~ B))

1 1 1 1 0 1 A=T, T˄T=T

1 1 0 1 1 0 A=T, T˄T=T

0 0 1 1 0 1 A=F, F˄T=F

0 0 0 1 1 0 A=F, F˄T=F

But if S is a label for (A˅ (B˅~B)), which is a tautology |1111|, then, if R is a tautology, then R→S is a tautology |1111| (as a formula).

(A ˅ (B ˅ ~ B))

1 1 1 1 0 1 A=T, T˅T=T

1 1 0 1 1 0 A=T, T˅T=T

0 1 1 1 0 1 A=F, F˅T=T

0 1 0 1 1 0 A=F, F˅T=T

R = (A v (B v ~B)) -> S = (A ^ (B v ~B)) 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 1 0 0 0 0 1 1

Date: Wed, 20 Mar 2013 04:59:09 +0000

To: timecoach@hotmail.com