Theorem excxor 1461

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 1457	. 2 ⊢ ((𝜑 ⊻ 𝜓) ↔ ¬ (𝜑 ↔ 𝜓))
2		xor 931	. 2 ⊢ (¬ (𝜑 ↔ 𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))
3		ancom 465	. . 3 ⊢ ((𝜓 ∧ ¬ 𝜑) ↔ (¬ 𝜑 ∧ 𝜓))
4	3	orbi2i 540	. 2 ⊢ (((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))
5	1, 2, 4	3bitri 285	1 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))

Syntax hints:  ¬ wn 3   ↔ wb 195   ∨ wo 382   ∧ wa 383   ⊻ wxo 1456

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 196  df-or 384  df-an 385  df-xor 1457

This theorem is referenced by:  f1omvdco2  17691  psgnunilem5  17737  or3or  37339