MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  xor Structured version   Visualization version   GIF version

Theorem xor 1014
Description: Two ways to express exclusive disjunction (df-xor 1511). 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 403 . . . 4 ((𝜑𝜓) ↔ ¬ (𝜑 ∧ ¬ 𝜓))
2 iman 403 . . . 4 ((𝜓𝜑) ↔ ¬ (𝜓 ∧ ¬ 𝜑))
31, 2anbi12i 628 . . 3 (((𝜑𝜓) ∧ (𝜓𝜑)) ↔ (¬ (𝜑 ∧ ¬ 𝜓) ∧ ¬ (𝜓 ∧ ¬ 𝜑)))
4 dfbi2 476 . . 3 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (𝜓𝜑)))
5 ioran 983 . . 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 397  wo 846
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 398  df-or 847
This theorem is referenced by:  pm5.24  1050  excxor  1516  elsymdif  4206  rpnnen2lem12  16067  ist0-3  22648  eliuniincex  43224  eliincex  43225  abnotataxb  45046  ldepslinc  46485
  Copyright terms: Public domain W3C validator