Theorem ifp1bi 39264

Description: Substitute the first element of conditional logical operator. (Contributed by RP, 20-Apr-2020.)

Assertion

Ref	Expression
ifp1bi	⊢ ((if-(𝜑, 𝜒, 𝜃) ↔ if-(𝜓, 𝜒, 𝜃)) ↔ ((((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜑 → 𝜓) ∨ (𝜃 → 𝜒))) ∧ (((𝜓 → 𝜑) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)))))

Proof of Theorem ifp1bi

Step	Hyp	Ref	Expression
1		dfbi2 467	. 2 ⊢ ((if-(𝜑, 𝜒, 𝜃) ↔ if-(𝜓, 𝜒, 𝜃)) ↔ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ∧ (if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃))))
2		ifpim1g 39263	. . . 4 ⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ∧ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃))))
3		ancom 453	. . . 4 ⊢ ((((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ∧ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃))) ↔ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))))
4	2, 3	bitri 267	. . 3 ⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))))
5		ifpim1g 39263	. . 3 ⊢ ((if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃)) ↔ (((𝜑 → 𝜓) ∨ (𝜃 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜒 → 𝜃))))
6	4, 5	anbi12i 617	. 2 ⊢ (((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ∧ (if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃))) ↔ ((((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))) ∧ (((𝜑 → 𝜓) ∨ (𝜃 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜒 → 𝜃)))))
7		an42 644	. 2 ⊢ (((((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))) ∧ (((𝜑 → 𝜓) ∨ (𝜃 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜒 → 𝜃)))) ↔ ((((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜑 → 𝜓) ∨ (𝜃 → 𝜒))) ∧ (((𝜓 → 𝜑) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)))))
8	1, 6, 7	3bitri 289	1 ⊢ ((if-(𝜑, 𝜒, 𝜃) ↔ if-(𝜓, 𝜒, 𝜃)) ↔ ((((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜑 → 𝜓) ∨ (𝜃 → 𝜒))) ∧ (((𝜓 → 𝜑) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)))))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 387   ∨ wo 833  if-wif 1043

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 199  df-an 388  df-or 834  df-ifp 1044

This theorem is referenced by: (None)