Let F1(A, B, C) = (A \[ \wedge \] \[ \sim \] B) \[ \vee \] [\[\sim\]C \[\wedge\] (A \[\vee\] B)] \[\vee\] \[\sim\] A and
F2(A, B) = (A \[\vee\] B) \[\vee\] (B \[ \to \] \[\sim\]A) be two logical expressions. Then :
Solutions · 1
Official solution
Truth table for F1 :
\[ \therefore \] F1 is not a tautology.
Truth table for F2 :
\[ \therefore \] F2 is a tautology.