Theorem xor 500

Description: Two ways to express "exclusive or". Theorem *5.22 of [WhiteheadRussell] p. 124.

Assertion

Ref	Expression
xor	⊢ (¬ (φ ↔ ψ) ↔ ((φ ∧ ¬ ψ) ∨ (ψ ∧ ¬ φ)))

Proof of Theorem xor

Step	Hyp	Ref	Expression
1		dfbi 499	. 2 ⊢ ((¬ φ ↔ ψ) ↔ ((¬ φ ∧ ψ) ∨ (¬ ¬ φ ∧ ¬ ψ)))
2		nbbn 498	. 2 ⊢ ((¬ φ ↔ ψ) ↔ ¬ (φ ↔ ψ))
3		ancom 333	. . . 4 ⊢ ((ψ ∧ ¬ φ) ↔ (¬ φ ∧ ψ))
4		pm4.13 142	. . . . 5 ⊢ (φ ↔ ¬ ¬ φ)
5	4	anbi1i 368	. . . 4 ⊢ ((φ ∧ ¬ ψ) ↔ (¬ ¬ φ ∧ ¬ ψ))
6	3, 5	orbi12i 216	. . 3 ⊢ (((ψ ∧ ¬ φ) ∨ (φ ∧ ¬ ψ)) ↔ ((¬ φ ∧ ψ) ∨ (¬ ¬ φ ∧ ¬ ψ)))
7		orcom 209	. . 3 ⊢ (((ψ ∧ ¬ φ) ∨ (φ ∧ ¬ ψ)) ↔ ((φ ∧ ¬ ψ) ∨ (ψ ∧ ¬ φ)))
8	6, 7	bitr3 153	. 2 ⊢ (((¬ φ ∧ ψ) ∨ (¬ ¬ φ ∧ ¬ ψ)) ↔ ((φ ∧ ¬ ψ) ∨ (ψ ∧ ¬ φ)))
9	1, 2, 8	3bitr3 156	1 ⊢ (¬ (φ ↔ ψ) ↔ ((φ ∧ ¬ ψ) ∨ (ψ ∧ ¬ φ)))

Syntax hints:  ¬ wn 1   ↔ wb 127   ∨ wo 195   ∧ wa 196

This theorem is referenced by:  biass 511

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

metamath.org