In: Advanced Math
FOR EAICH PAIR OF PROPOSITIONS P AND Q STATE WHETHER ON NOT p=q
p=(s→(p ∧¬r)) ∧ ((p→(r ∨ q)) ∧ s), Q=p ∨ t
Truth tabl
P | Q | |||||||||||
p | q | r | s | t | ~r | p^~r | s->(p^~r) | rVq | p->(rVq) | (p->(rV q))^s | (s->(p^~r))^((p->(rVq))^s) | pVt |
T | T | T | T | T | F | T | T | T | T | T | T | T |
T | T | T | T | F | F | F | F | T | T | T | F | T |
T | T | T | F | T | F | T | T | T | T | F | F | T |
T | T | T | F | F | F | F | T | T | T | F | F | T |
T | T | F | T | T | T | T | T | T | T | T | T | T |
T | T | F | T | F | T | F | F | T | T | T | F | T |
T | T | F | F | T | T | T | T | T | T | F | F | T |
T | T | F | F | F | T | F | T | T | T | F | F | T |
T | F | T | T | T | F | T | T | T | T | T | T | T |
T | F | T | T | F | F | F | F | T | T | T | F | T |
T | F | T | F | T | F | T | T | T | T | F | F | T |
T | F | T | F | F | F | F | T | T | T | F | F | T |
T | F | F | T | T | T | T | T | F | F | F | F | T |
T | F | F | T | F | T | F | F | F | F | F | F | T |
T | F | F | F | T | T | T | T | F | F | F | F | T |
T | F | F | F | F | T | F | T | F | F | F | F | T |
F | T | T | T | T | F | F | F | T | T | T | F | T |
F | T | T | T | F | F | F | F | T | T | T | F | F |
F | T | T | F | T | F | F | T | T | T | F | F | T |
F | T | T | F | F | F | F | T | T | T | F | F | F |
F | T | F | T | T | T | F | F | T | T | T | F | T |
F | T | F | T | F | T | F | F | T | T | T | F | F |
F | T | F | F | T | T | F | T | T | T | F | F | T |
F | T | F | F | F | T | F | T | T | T | F | F | F |
F | F | T | T | T | F | F | F | T | T | T | F | T |
F | F | T | T | F | F | F | F | T | T | T | F | F |
F | F | T | F | T | F | F | T | T | T | F | F | T |
F | F | T | F | F | F | F | T | T | T | F | F | F |
F | F | F | T | T | T | F | F | F | T | T | F | T |
F | F | F | T | F | T | F | F | F | T | T | F | F |
F | F | F | F | T | T | F | T | F | T | F | F | T |
F | F | F | F | F | T | F | T | F | T | F | F | F |
fBy analyzing truth table we can conclude that P != Q.