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Theorem xor 1010
Description: Two ways to express exclusive disjunction (df-xor 1498). 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 626 . . 3 (((𝜑𝜓) ∧ (𝜓𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
4 dfbi2 475 . . 3 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (𝜓𝜑)))
5 ioran 979 . . 3 (¬ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
63, 4, 53bitr4ri 305 . 2 (¬ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)) ↔ (𝜑𝜓))
76con1bii 358 1 (¬ (𝜑𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (𝜓 ∧ ¬ 𝜑)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 207  wa 396  wo 843
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 208  df-an 397  df-or 844
This theorem is referenced by:  pm5.24  1044  excxor  1502  elsymdif  4227  rpnnen2lem12  15570  ist0-3  21869  eliuniincex  41238  eliincex  41239  abnotataxb  43015  ldepslinc  44393
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