Theorem excxor 1368

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 1367	. . 3 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)))
2		andi 808	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)) ↔ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ∨ ((𝜑 ∨ 𝜓) ∧ ¬ 𝜓)))
3		orcom 718	. . . . 5 ⊢ (((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝜑) ∨ (𝜓 ∧ ¬ 𝜑)))
4		pm3.24 683	. . . . . 6 ⊢ ¬ (𝜑 ∧ ¬ 𝜑)
5	4	biorfi 736	. . . . 5 ⊢ ((𝜓 ∧ ¬ 𝜑) ↔ ((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜑)))
6		andir 809	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ↔ ((𝜑 ∧ ¬ 𝜑) ∨ (𝜓 ∧ ¬ 𝜑)))
7	3, 5, 6	3bitr4ri 212	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ↔ (𝜓 ∧ ¬ 𝜑))
8		pm5.61 784	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ 𝜓) ↔ (𝜑 ∧ ¬ 𝜓))
9	7, 8	orbi12i 754	. . 3 ⊢ ((((𝜑 ∨ 𝜓) ∧ ¬ 𝜑) ∨ ((𝜑 ∨ 𝜓) ∧ ¬ 𝜓)) ↔ ((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜓)))
10	1, 2, 9	3bitri 205	. 2 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜓)))
11		orcom 718	. 2 ⊢ (((𝜓 ∧ ¬ 𝜑) ∨ (𝜑 ∧ ¬ 𝜓)) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))
12		ancom 264	. . 3 ⊢ ((𝜓 ∧ ¬ 𝜑) ↔ (¬ 𝜑 ∧ 𝜓))
13	12	orbi2i 752	. 2 ⊢ (((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))
14	10, 11, 13	3bitri 205	1 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))

Syntax hints:  ¬ wn 3   ∧ wa 103   ↔ wb 104   ∨ wo 698   ⊻ wxo 1365

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699

This theorem depends on definitions:  df-bi 116  df-xor 1366

This theorem is referenced by:  xordc  1382  symdifxor  3388