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Theorem xor2 675
Description: Two ways to express "exclusive or."
Assertion
Ref Expression
xor2 (¬ (φψ) ↔ ((φ ψ) ¬ (φ ψ)))

Proof of Theorem xor2
StepHypRef Expression
1 xor 673 . 2 (¬ (φψ) ↔ ((φ ¬ ψ) (ψ ¬ φ)))
2 ioran 306 . . 3 (¬ ((φ ψ) φ ¬ ψ)) ↔ (¬ (φ ψ) ¬ (¬ φ ¬ ψ)))
3 pm5.24 674 . . 3 (¬ ((φ ψ) φ ¬ ψ)) ↔ ((φ ¬ ψ) (ψ ¬ φ)))
4 oran 312 . . . . 5 ((φ ψ) ↔ ¬ (¬ φ ¬ ψ))
54anbi2i 482 . . . 4 ((¬ (φ ψ) (φ ψ)) ↔ (¬ (φ ψ) ¬ (¬ φ ¬ ψ)))
6 ancom 437 . . . 4 ((¬ (φ ψ) (φ ψ)) ↔ ((φ ψ) ¬ (φ ψ)))
75, 6bitr3 175 . . 3 ((¬ (φ ψ) ¬ (¬ φ ¬ ψ)) ↔ ((φ ψ) ¬ (φ ψ)))
82, 3, 73bitr3 181 . 2 (((φ ¬ ψ) (ψ ¬ φ)) ↔ ((φ ψ) ¬ (φ ψ)))
91, 8bitr 173 1 (¬ (φψ) ↔ ((φ ψ) ¬ (φ ψ)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 2   ↔ wb 146   wo 222   wa 223
This theorem is referenced by:  nmogtmnf 8429  nmopgtmnf 9790
This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7
This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225
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