Theorem excxor 1389

Description: This tautology shows that xor is really exclusive. (Contributed by FL, 22-Nov-2010.) (Proof rewritten by Jim Kingdon, 5-May-2018.)

Assertion

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

Proof of Theorem excxor

Step	Hyp	Ref	Expression
1		xoranor 1388	. . 3 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)))
2		andi 819	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)) ↔ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ∨ ((𝜑 ∨ 𝜓) ∧ ¬ 𝜓)))
3		orcom 729	. . . . 5 ⊢ (((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝜑) ∨ (𝜓 ∧ ¬ 𝜑)))
4		pm3.24 694	. . . . . 6 ⊢ ¬ (𝜑 ∧ ¬ 𝜑)
5	4	biorfi 747	. . . . 5 ⊢ ((𝜓 ∧ ¬ 𝜑) ↔ ((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜑)))
6		andir 820	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ↔ ((𝜑 ∧ ¬ 𝜑) ∨ (𝜓 ∧ ¬ 𝜑)))
7	3, 5, 6	3bitr4ri 213	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ↔ (𝜓 ∧ ¬ 𝜑))
8		pm5.61 795	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜓) ↔ (𝜑 ∧ ¬ 𝜓))
9	7, 8	orbi12i 765	. . 3 ⊢ ((((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ∨ ((𝜑 ∨ 𝜓) ∧ ¬ 𝜓)) ↔ ((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜓)))
10	1, 2, 9	3bitri 206	. 2 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜓)))
11		orcom 729	. 2 ⊢ (((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜓)) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))
12		ancom 266	. . 3 ⊢ ((𝜓 ∧ ¬ 𝜑) ↔ (¬ 𝜑 ∧ 𝜓))
13	12	orbi2i 763	. 2 ⊢ (((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))
14	10, 11, 13	3bitri 206	1 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))

Syntax hints:  ¬ wn 3   ∧ wa 104   ↔ wb 105   ∨ wo 709   ⊻ wxo 1386

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710

This theorem depends on definitions:  df-bi 117  df-xor 1387

This theorem is referenced by:  xordc  1403  symdifxor  3425