Theorem ifpim1g 44486

Description: Implication of conditional logical operators. (Contributed by RP, 18-Apr-2020.)

Assertion

Ref	Expression
ifpim1g	⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ∧ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃))))

Proof of Theorem ifpim1g

Step	Hyp	Ref	Expression
1		ifpim123g 44485	. 2 ⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ ((((𝜑 → ¬ 𝜓) ∨ (𝜒 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))) ∧ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((¬ 𝜓 → 𝜑) ∨ (𝜃 → 𝜃)))))
2		id 23	. . . . . 6 ⊢ (𝜒 → 𝜒)
3	2	olci 880	. . . . 5 ⊢ ((𝜑 → ¬ 𝜓) ∨ (𝜒 → 𝜒))
4	3	biantrur 540	. . . 4 ⊢ (((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ↔ (((𝜑 → ¬ 𝜓) ∨ (𝜒 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))))
5	4	bicomi 227	. . 3 ⊢ ((((𝜑 → ¬ 𝜓) ∨ (𝜒 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))) ↔ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)))
6		id 23	. . . . . 6 ⊢ (𝜃 → 𝜃)
7	6	olci 880	. . . . 5 ⊢ ((¬ 𝜓 → 𝜑) ∨ (𝜃 → 𝜃))
8	7	biantru 539	. . . 4 ⊢ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ↔ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((¬ 𝜓 → 𝜑) ∨ (𝜃 → 𝜃))))
9	8	bicomi 227	. . 3 ⊢ ((((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((¬ 𝜓 → 𝜑) ∨ (𝜃 → 𝜃))) ↔ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)))
10	5, 9	anbi12i 640	. 2 ⊢ (((((𝜑 → ¬ 𝜓) ∨ (𝜒 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))) ∧ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((¬ 𝜓 → 𝜑) ∨ (𝜃 → 𝜃)))) ↔ (((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ∧ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃))))
11	1, 10	bitri 278	1 ⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ∧ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃))))

This proof depends on syntax axioms:  ¬ wn 3   → 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:  ifp1bi  44487