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Theorem xor 1012
Description: Two ways to express exclusive disjunction (df-xor 1507). 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 (¬ (𝜑𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))

Proof of Theorem xor
StepHypRef Expression
1 iman 402 . . . 4 ((𝜑𝜓) ↔ ¬ (𝜑 ∧ ¬ 𝜓))
2 iman 402 . . . 4 ((𝜓𝜑) ↔ ¬ (𝜓 ∧ ¬ 𝜑))
31, 2anbi12i 627 . . 3 (((𝜑𝜓) ∧ (𝜓𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
4 dfbi2 475 . . 3 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (𝜓𝜑)))
5 ioran 981 . . 3 (¬ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
63, 4, 53bitr4ri 304 . 2 (¬ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ (𝜑𝜓))
76con1bii 357 1 (¬ (𝜑𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 396  wo 844
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 206  df-an 397  df-or 845
This theorem is referenced by:  pm5.24  1048  excxor  1512  elsymdif  4181  rpnnen2lem12  15934  ist0-3  22496  eliuniincex  42659  eliincex  42660  abnotataxb  44411  ldepslinc  45850
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