Theorem nanorxor 37526

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 1440	. 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
2		xor2 1462	. . . 4 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)))
3	2	rbaibr 944	. . 3 ⊢ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
4	2	bibi2i 326	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) ↔ ((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓))))
5		pm4.71 660	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) → ¬ (𝜑 ∧ 𝜓)) ↔ ((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓))))
6		simpl 472	. . . . . . . 8 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
7	6	orcd 406	. . . . . . 7 ⊢ ((𝜑 ∧ 𝜓) → (𝜑 ∨ 𝜓))
8	7	con3i 149	. . . . . 6 ⊢ (¬ (𝜑 ∨ 𝜓) → ¬ (𝜑 ∧ 𝜓))
9		id 22	. . . . . 6 ⊢ (¬ (𝜑 ∧ 𝜓) → ¬ (𝜑 ∧ 𝜓))
10	8, 9	ja 172	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) → ¬ (𝜑 ∧ 𝜓)) → ¬ (𝜑 ∧ 𝜓))
11	5, 10	sylbir 224	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓))) → ¬ (𝜑 ∧ 𝜓))
12	4, 11	sylbi 206	. . 3 ⊢ (((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)) → ¬ (𝜑 ∧ 𝜓))
13	3, 12	impbii 198	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))
14	1, 13	bitri 263	1 ⊢ ((𝜑 ⊼ 𝜓) ↔ ((𝜑 ∨ 𝜓) ↔ (𝜑 ⊻ 𝜓)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∨ wo 382   ∧ wa 383   ⊼ wnan 1439   ⊻ wxo 1456

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 196  df-or 384  df-an 385  df-nan 1440  df-xor 1457

This theorem is referenced by:  undisjrab  37527