Theorem excxor 1309

Description: This tautology shows that xor is really exclusive. (Contributed by FL, 22-Nov-2010.)

Assertion

Ref	Expression
excxor	⊢ ((φ ⊻ ψ) ↔ ((φ ∧ ¬ ψ) ∨ (¬ φ ∧ ψ)))

Proof of Theorem excxor

Step	Hyp	Ref	Expression
1		df-xor 1305	. 2 ⊢ ((φ ⊻ ψ) ↔ ¬ (φ ↔ ψ))
2		xor 861	. 2 ⊢ (¬ (φ ↔ ψ) ↔ ((φ ∧ ¬ ψ) ∨ (ψ ∧ ¬ φ)))
3		ancom 437	. . 3 ⊢ ((ψ ∧ ¬ φ) ↔ (¬ φ ∧ ψ))
4	3	orbi2i 505	. 2 ⊢ (((φ ∧ ¬ ψ) ∨ (ψ ∧ ¬ φ)) ↔ ((φ ∧ ¬ ψ) ∨ (¬ φ ∧ ψ)))
5	1, 2, 4	3bitri 262	1 ⊢ ((φ ⊻ ψ) ↔ ((φ ∧ ¬ ψ) ∨ (¬ φ ∧ ψ)))

Syntax hints:  ¬ wn 3   ↔ wb 176   ∨ wo 357   ∧ wa 358   ⊻ wxo 1304

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-xor 1305

This theorem is referenced by: (None)