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Theorem falxorfal 1401
 Description: A ⊻ identity. (Contributed by David A. Wheeler, 2-Mar-2018.)
Assertion
Ref Expression
falxorfal ((⊥ ⊻ ⊥) ↔ ⊥)

Proof of Theorem falxorfal
StepHypRef Expression
1 df-xor 1355 . 2 ((⊥ ⊻ ⊥) ↔ ((⊥ ∨ ⊥) ∧ ¬ (⊥ ∧ ⊥)))
2 oridm 747 . . 3 ((⊥ ∨ ⊥) ↔ ⊥)
3 notfal 1393 . . . 4 (¬ ⊥ ↔ ⊤)
4 anidm 394 . . . 4 ((⊥ ∧ ⊥) ↔ ⊥)
53, 4xchnxbir 671 . . 3 (¬ (⊥ ∧ ⊥) ↔ ⊤)
62, 5anbi12i 456 . 2 (((⊥ ∨ ⊥) ∧ ¬ (⊥ ∧ ⊥)) ↔ (⊥ ∧ ⊤))
7 falantru 1382 . 2 ((⊥ ∧ ⊤) ↔ ⊥)
81, 6, 73bitri 205 1 ((⊥ ⊻ ⊥) ↔ ⊥)
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   ∧ wa 103   ↔ wb 104   ∨ wo 698  ⊤wtru 1333  ⊥wfal 1337   ⊻ wxo 1354 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699 This theorem depends on definitions:  df-bi 116  df-tru 1335  df-fal 1338  df-xor 1355 This theorem is referenced by: (None)
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