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Theorem falxorfal 1471
Description: A identity. (Contributed by David A. Wheeler, 2-Mar-2018.)
Assertion
Ref Expression
falxorfal ((⊥ ⊻ ⊥) ↔ ⊥)

Proof of Theorem falxorfal
StepHypRef Expression
1 df-xor 1425 . 2 ((⊥ ⊻ ⊥) ↔ ((⊥ ∨ ⊥) ∧ ¬ (⊥ ∧ ⊥)))
2 oridm 769 . . 3 ((⊥ ∨ ⊥) ↔ ⊥)
3 notfal 1463 . . . 4 (¬ ⊥ ↔ ⊤)
4 anidm 400 . . . 4 ((⊥ ∧ ⊥) ↔ ⊥)
53, 4xchnxbir 692 . . 3 (¬ (⊥ ∧ ⊥) ↔ ⊤)
62, 5anbi12i 464 . 2 (((⊥ ∨ ⊥) ∧ ¬ (⊥ ∧ ⊥)) ↔ (⊥ ∧ ⊤))
7 falantru 1452 . 2 ((⊥ ∧ ⊤) ↔ ⊥)
81, 6, 73bitri 206 1 ((⊥ ⊻ ⊥) ↔ ⊥)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wa 104  wb 105  wo 720  wtru 1403  wfal 1407  wxo 1424
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721
This theorem depends on definitions:  df-bi 117  df-tru 1405  df-fal 1408  df-xor 1425
This theorem is referenced by: (None)
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