Theorem ifpid1g 44453

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 44450	. 2 ⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ∧ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒)))))
2		ancom 466	. 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		pm4.25 919	. . . . 5 ⊢ (𝜑 ↔ (𝜑 ∨ 𝜑))
4	3	imbi2i 339	. . . 4 ⊢ ((𝜒 → 𝜑) ↔ (𝜒 → (𝜑 ∨ 𝜑)))
5		orc 881	. . . . 5 ⊢ (𝜑 → (𝜑 ∨ 𝜒))
6	5	biantru 539	. . . 4 ⊢ ((𝜒 → (𝜑 ∨ 𝜑)) ↔ ((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))))
7	4, 6	bitr2i 279	. . 3 ⊢ (((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ↔ (𝜒 → 𝜑))
8		pm4.24 574	. . . . 5 ⊢ (𝜑 ↔ (𝜑 ∧ 𝜑))
9	8	imbi1i 352	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ ((𝜑 ∧ 𝜑) → 𝜓))
10		simpl 488	. . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
11	10	biantrur 540	. . . 4 ⊢ (((𝜑 ∧ 𝜑) → 𝜓) ↔ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)))
12	9, 11	bitr2i 279	. . 3 ⊢ ((((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓)) ↔ (𝜑 → 𝜓))
13	7, 12	anbi12i 640	. 2 ⊢ ((((𝜒 → (𝜑 ∨ 𝜑)) ∧ (𝜑 → (𝜑 ∨ 𝜒))) ∧ (((𝜑 ∧ 𝜓) → 𝜑) ∧ ((𝜑 ∧ 𝜑) → 𝜓))) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))
14	1, 2, 13	3bitri 300	1 ⊢ ((𝜑 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((𝜒 → 𝜑) ∧ (𝜑 → 𝜓)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861  if-wif 1078

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ifp 1079

This theorem is used by: (None)