Theorem ifpid1g 36858

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

Assertion

Ref	Expression
ifpid1g	⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))

Proof of Theorem ifpid1g

Step	Hyp	Ref	Expression
1		ifpidg 36855	. 2 ⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ∧ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒)))))
2		ancom 465	. 2 ⊢ (((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ∧ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒)))) ↔ (((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ∧ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓))))
3		pm4.25 536	. . . . 5 ⊢ (𝜑 ↔ (𝜑 ∨ 𝜑))
4	3	imbi2i 325	. . . 4 ⊢ ((𝜒 → 𝜑) ↔ (𝜒 → (𝜑 ∨ 𝜑)))
5		orc 399	. . . . 5 ⊢ (𝜑 → (𝜑 ∨ 𝜒))
6	5	biantru 525	. . . 4 ⊢ ((𝜒 → (𝜑 ∨ 𝜑)) ↔ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))))
7	4, 6	bitr2i 264	. . 3 ⊢ (((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ↔ (𝜒 → 𝜑))
8		pm4.24 673	. . . . 5 ⊢ (𝜑 ↔ (𝜑 ∧ 𝜑))
9	8	imbi1i 338	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ ((𝜑 ∧ 𝜑) → 𝜓))
10		simpl 472	. . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
11	10	biantrur 526	. . . 4 ⊢ (((𝜑 ∧ 𝜑) → 𝜓) ↔ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)))
12	9, 11	bitr2i 264	. . 3 ⊢ ((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ↔ (𝜑 → 𝜓))
13	7, 12	anbi12i 729	. 2 ⊢ ((((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ∧ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓))) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))
14	1, 2, 13	3bitri 285	1 ⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))

Syntax hints:   → wi 4   ↔ wb 195   ∨ wo 382   ∧ wa 383  if-wif 1006

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

This theorem is referenced by: (None)