Theorem xor 1032

Description: Two ways to express exclusive disjunction (df-xor 1542). Theorem *5.22 of [WhiteheadRussell] p. 124. (Contributed by NM, 3-Jan-2005.) (Proof shortened by Wolf Lammen, 22-Jan-2013.)

Assertion

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

Proof of Theorem xor

Step	Hyp	Ref	Expression
1		iman 407	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ ¬ (𝜑 ∧ ¬ 𝜓))
2		iman 407	. . . 4 ⊢ ((𝜓 → 𝜑) ↔ ¬ (𝜓 ∧ ¬ 𝜑))
3	1, 2	anbi12i 640	. . 3 ⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
4		dfbi2 480	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
5		ioran 999	. . 3 ⊢ (¬ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
6	3, 4, 5	3bitr4ri 307	. 2 ⊢ (¬ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ (𝜑 ↔ 𝜓))
7	6	con1bii 359	1 ⊢ (¬ (𝜑 ↔ 𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862

This theorem is used by:  pm5.24  1066  excxor  1546  elsymdif  4214  rpnnen2lem12  16306  ist0-3  23539  eliuniincex  45868  eliincex  45869  abnotataxb  47694  ldepslinc  49330