Theorem ifpid1g 41063

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 41060	. 2 ⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ∧ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒)))))
2		ancom 460	. 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		pm4.25 902	. . . . 5 ⊢ (𝜑 ↔ (𝜑 ∨ 𝜑))
4	3	imbi2i 335	. . . 4 ⊢ ((𝜒 → 𝜑) ↔ (𝜒 → (𝜑 ∨ 𝜑)))
5		orc 863	. . . . 5 ⊢ (𝜑 → (𝜑 ∨ 𝜒))
6	5	biantru 529	. . . 4 ⊢ ((𝜒 → (𝜑 ∨ 𝜑)) ↔ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))))
7	4, 6	bitr2i 275	. . 3 ⊢ (((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ↔ (𝜒 → 𝜑))
8		pm4.24 563	. . . . 5 ⊢ (𝜑 ↔ (𝜑 ∧ 𝜑))
9	8	imbi1i 349	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ ((𝜑 ∧ 𝜑) → 𝜓))
10		simpl 482	. . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
11	10	biantrur 530	. . . 4 ⊢ (((𝜑 ∧ 𝜑) → 𝜓) ↔ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)))
12	9, 11	bitr2i 275	. . 3 ⊢ ((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ↔ (𝜑 → 𝜓))
13	7, 12	anbi12i 626	. 2 ⊢ ((((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ∧ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓))) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))
14	1, 2, 13	3bitri 296	1 ⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843  if-wif 1059

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 206  df-an 396  df-or 844  df-ifp 1060

This theorem is referenced by: (None)