It's a cinch to prove it with truth tables, and murder to prove it by
argument forms. Since all the argument forms are proved by truth tables,
your teacher should allow a truth table proof.
[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
---------------------------
T ≡ (T ≡ T) ≡ (T ≡ T) ≡ T
T ≡ (T ≡ F) ≡ (T ≡ T) ≡ F
T ≡ (F ≡ T) ≡ (T ≡ F) ≡ T
T ≡ (F ≡ F) ≡ (T ≡ F) ≡ F
F ≡ (T ≡ T) ≡ (F ≡ T) ≡ T
F ≡ (T ≡ F) ≡ (F ≡ T) ≡ F
F ≡ (F ≡ T) ≡ (F ≡ F) ≡ T
F ≡ (F ≡ F) ≡ (F ≡ F) ≡ F
Do equivalence inside the parentheses
If the values are the same on both sides
of the ≡, put T, otherwise put F.
Then erase what you used to get what
you just put down.
[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
-----------------------------
[T ≡ T ] ≡ [ T ≡ T]
[T ≡ F ] ≡ [ T ≡ F]
[T ≡ F ] ≡ [ F ≡ T]
[T ≡ T ] ≡ [ F ≡ F]
[F ≡ T ] ≡ [ F ≡ T]
[F ≡ F ] ≡ [ F ≡ F]
[F ≡ F ] ≡ [ T ≡ T]
[F ≡ T ] ≡ [ T ≡ F]
Now do equivalence inside the brackets
As before, if the values are the same
on both sides of the ≡, put T, otherwise
put F. Then erase what you used to get
what you just put down.
[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
-----------------------------
T ≡ T
F ≡ F
F ≡ F
T ≡ T
F ≡ F
T ≡ T
T ≡ T
F ≡ F
Now do the final equivalence inside the
brackets As before, if the values are the
same on both sides of the ≡, put T, otherwise
put F. Then erase what you used to get
what you just put down.
[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
-----------------------------
T
T
T
T
T
T
T
T
Since we end up with all T's, that proves
that [A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C] is a
tautology.
Edwin
