Theorem falxorfal 1471

Description: A ⊻ identity. (Contributed by David A. Wheeler, 2-Mar-2018.)

Assertion

Ref	Expression
falxorfal	⊢ ((⊥ ⊻ ⊥) ↔ ⊥)

Proof of Theorem falxorfal

Step	Hyp	Ref	Expression
1		df-xor 1425	. 2 ⊢ ((⊥ ⊻ ⊥) ↔ ((⊥ ∨ ⊥) ∧ ¬ (⊥ ∧ ⊥)))
2		oridm 769	. . 3 ⊢ ((⊥ ∨ ⊥) ↔ ⊥)
3		notfal 1463	. . . 4 ⊢ (¬ ⊥ ↔ ⊤)
4		anidm 400	. . . 4 ⊢ ((⊥ ∧ ⊥) ↔ ⊥)
5	3, 4	xchnxbir 692	. . 3 ⊢ (¬ (⊥ ∧ ⊥) ↔ ⊤)
6	2, 5	anbi12i 464	. 2 ⊢ (((⊥ ∨ ⊥) ∧ ¬ (⊥ ∧ ⊥)) ↔ (⊥ ∧ ⊤))
7		falantru 1452	. 2 ⊢ ((⊥ ∧ ⊤) ↔ ⊥)
8	1, 6, 7	3bitri 206	1 ⊢ ((⊥ ⊻ ⊥) ↔ ⊥)

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 104   ↔ wb 105   ∨ wo 720  ⊤wtru 1403  ⊥wfal 1407   ⊻ wxo 1424

This proof depends on 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 proof depends on definitions:  df-bi 117  df-tru 1405  df-fal 1408  df-xor 1425

This theorem is used by: (None)