I came across following problem:
Is following tautology:
$((A ∧ B) → C) ≡ ((A → C) ∨ (B → C))$
I solved it like this:
$(A\wedge B)\rightarrow C$
$=\neg(A\wedge B)\vee C$
$=(\neg A\vee \neg B) \vee C$
$=\neg A\vee \neg B \vee C$
$=\neg A\vee \neg B \vee C \vee C$ $\hspace{15pt}(\because C\vee C=C)$
$=(\neg A\vee C)\vee(\neg B\vee C )$
$=(A\rightarrow C)\vee(B\rightarrow C)$
So I concluded that it is a tautology.
But when I checked the answer it was given that it is not a tautology. So I checked wolframalpha and it also seem to give different truth tables for the two:
$(A\rightarrow C)\vee(B\rightarrow C)$:
Where did I make mistake?
Best Answer
The truth tables are the same, the variables are simply labelled in a different order in your tables.
Table 1 has $A$, $B$ then $C$, table 2 has $A$, $C$ then $B$.
Look: