Theorem ifpdfbi 36837

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

Assertion

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

Proof of Theorem ifpdfbi

Step	Hyp	Ref	Expression
1		dfbi2 658	. 2 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2		ifpim1 36832	. . . . 5 ⊢ ((𝜑 → 𝜓) ↔ if-(¬ 𝜑, ⊤, 𝜓))
3		ifpn 1016	. . . . 5 ⊢ (if-(𝜑, 𝜓, ⊤) ↔ if-(¬ 𝜑, ⊤, 𝜓))
4	2, 3	bitr4i 266	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ if-(𝜑, 𝜓, ⊤))
5		ifpim2 36835	. . . 4 ⊢ ((𝜓 → 𝜑) ↔ if-(𝜑, ⊤, ¬ 𝜓))
6	4, 5	anbi12i 729	. . 3 ⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ (if-(𝜑, 𝜓, ⊤) ∧ if-(𝜑, ⊤, ¬ 𝜓)))
7		ifpan23 36823	. . . 4 ⊢ ((if-(𝜑, 𝜓, ⊤) ∧ if-(𝜑, ⊤, ¬ 𝜓)) ↔ if-(𝜑, (𝜓 ∧ ⊤), (⊤ ∧ ¬ 𝜓)))
8		ancom 465	. . . . . 6 ⊢ ((𝜓 ∧ ⊤) ↔ (⊤ ∧ 𝜓))
9		truan 1492	. . . . . 6 ⊢ ((⊤ ∧ 𝜓) ↔ 𝜓)
10	8, 9	bitri 263	. . . . 5 ⊢ ((𝜓 ∧ ⊤) ↔ 𝜓)
11		truan 1492	. . . . 5 ⊢ ((⊤ ∧ ¬ 𝜓) ↔ ¬ 𝜓)
12		ifpbi23 36836	. . . . 5 ⊢ ((((𝜓 ∧ ⊤) ↔ 𝜓) ∧ ((⊤ ∧ ¬ 𝜓) ↔ ¬ 𝜓)) → (if-(𝜑, (𝜓 ∧ ⊤), (⊤ ∧ ¬ 𝜓)) ↔ if-(𝜑, 𝜓, ¬ 𝜓)))
13	10, 11, 12	mp2an 704	. . . 4 ⊢ (if-(𝜑, (𝜓 ∧ ⊤), (⊤ ∧ ¬ 𝜓)) ↔ if-(𝜑, 𝜓, ¬ 𝜓))
14	7, 13	bitri 263	. . 3 ⊢ ((if-(𝜑, 𝜓, ⊤) ∧ if-(𝜑, ⊤, ¬ 𝜓)) ↔ if-(𝜑, 𝜓, ¬ 𝜓))
15	6, 14	bitri 263	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ if-(𝜑, 𝜓, ¬ 𝜓))
16	1, 15	bitri 263	1 ⊢ ((𝜑 ↔ 𝜓) ↔ if-(𝜑, 𝜓, ¬ 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∧ wa 383  if-wif 1006  ⊤wtru 1476

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-tru 1478

This theorem is referenced by:  ifpbiidcor  36838  ifpbicor  36839