Theorem ifpdfxor 36851

Description: Define xor as conditional logic operator. (Contributed by RP, 20-Apr-2020.)

Assertion

Ref	Expression
ifpdfxor	⊢ ((𝜑 ⊻ 𝜓) ↔ if-(𝜑, ¬ 𝜓, 𝜓))

Proof of Theorem ifpdfxor

Step	Hyp	Ref	Expression
1		xor2 1462	. 2 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)))
2		ifpdfor 36828	. . 3 ⊢ ((𝜑 ∨ 𝜓) ↔ if-(𝜑, ⊤, 𝜓))
3		ifpnot23 36842	. . . 4 ⊢ (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ¬ ⊥))
4		ifpdfan 36829	. . . 4 ⊢ ((𝜑 ∧ 𝜓) ↔ if-(𝜑, 𝜓, ⊥))
5	3, 4	xchnxbir 322	. . 3 ⊢ (¬ (𝜑 ∧ 𝜓) ↔ if-(𝜑, ¬ 𝜓, ¬ ⊥))
6	2, 5	anbi12i 729	. 2 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)) ↔ (if-(𝜑, ⊤, 𝜓) ∧ if-(𝜑, ¬ 𝜓, ¬ ⊥)))
7		ifpan23 36823	. . 3 ⊢ ((if-(𝜑, ⊤, 𝜓) ∧ if-(𝜑, ¬ 𝜓, ¬ ⊥)) ↔ if-(𝜑, (⊤ ∧ ¬ 𝜓), (𝜓 ∧ ¬ ⊥)))
8		truan 1492	. . . 4 ⊢ ((⊤ ∧ ¬ 𝜓) ↔ ¬ 𝜓)
9		fal 1482	. . . . . 6 ⊢ ¬ ⊥
10	9	biantru 525	. . . . 5 ⊢ (𝜓 ↔ (𝜓 ∧ ¬ ⊥))
11	10	bicomi 213	. . . 4 ⊢ ((𝜓 ∧ ¬ ⊥) ↔ 𝜓)
12		ifpbi23 36836	. . . 4 ⊢ ((((⊤ ∧ ¬ 𝜓) ↔ ¬ 𝜓) ∧ ((𝜓 ∧ ¬ ⊥) ↔ 𝜓)) → (if-(𝜑, (⊤ ∧ ¬ 𝜓), (𝜓 ∧ ¬ ⊥)) ↔ if-(𝜑, ¬ 𝜓, 𝜓)))
13	8, 11, 12	mp2an 704	. . 3 ⊢ (if-(𝜑, (⊤ ∧ ¬ 𝜓), (𝜓 ∧ ¬ ⊥)) ↔ if-(𝜑, ¬ 𝜓, 𝜓))
14	7, 13	bitri 263	. 2 ⊢ ((if-(𝜑, ⊤, 𝜓) ∧ if-(𝜑, ¬ 𝜓, ¬ ⊥)) ↔ if-(𝜑, ¬ 𝜓, 𝜓))
15	1, 6, 14	3bitri 285	1 ⊢ ((𝜑 ⊻ 𝜓) ↔ if-(𝜑, ¬ 𝜓, 𝜓))

Syntax hints:  ¬ wn 3   ↔ wb 195   ∨ wo 382   ∧ wa 383  if-wif 1006   ⊻ wxo 1456  ⊤wtru 1476  ⊥wfal 1480

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-ifp 1007  df-xor 1457  df-tru 1478  df-fal 1481

This theorem is referenced by: (None)