Theorem nanorxor 40630

Description: 'nand' is equivalent to the equivalence of inclusive and exclusive or. (Contributed by Steve Rodriguez, 28-Feb-2020.)

Assertion

Ref	Expression
nanorxor	⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))

Proof of Theorem nanorxor

Step	Hyp	Ref	Expression
1		df-nan 1481	. 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
2		xor2 1506	. . . 4 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)))
3	2	rbaibr 540	. . 3 ⊢ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2	bibi2i 340	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) ↔ ((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓))))
5		pm4.71 560	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) → ¬ (𝜑 ∧ 𝜓)) ↔ ((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓))))
6		simpl 485	. . . . . . . 8 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
7	6	orcd 869	. . . . . . 7 ⊢ ((𝜑 ∧ 𝜓) → (𝜑 ∨ 𝜓))
8	7	con3i 157	. . . . . 6 ⊢ (¬ (𝜑 ∨ 𝜓) → ¬ (𝜑 ∧ 𝜓))
9		id 22	. . . . . 6 ⊢ (¬ (𝜑 ∧ 𝜓) → ¬ (𝜑 ∧ 𝜓))
10	8, 9	ja 188	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) → ¬ (𝜑 ∧ 𝜓)) → ¬ (𝜑 ∧ 𝜓))
11	5, 10	sylbir 237	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓))) → ¬ (𝜑 ∧ 𝜓))
12	4, 11	sylbi 219	. . 3 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) → ¬ (𝜑 ∧ 𝜓))
13	3, 12	impbii 211	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
14	1, 13	bitri 277	1 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398   ∨ wo 843   ⊼ wnan 1480   ⊻ wxo 1500

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 209  df-an 399  df-or 844  df-nan 1481  df-xor 1501

This theorem is referenced by:  undisjrab  40631