Theorem ifp1bi 40794

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 478	. 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 40793	. . . 4 ⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜓 → 𝜑) ∨ (𝜃 → 𝜒)) ∧ ((𝜑 → 𝜓) ∨ (𝜒 → 𝜃))))
3	2	biancomi 466	. . 3 ⊢ ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜑 → 𝜓) ∨ (𝜒 → 𝜃)) ∧ ((𝜓 → 𝜑) ∨ (𝜃 → 𝜒))))
4		ifpim1g 40793	. . 3 ⊢ ((if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃)) ↔ (((𝜑 → 𝜓) ∨ (𝜃 → 𝜒)) ∧ ((𝜓 → 𝜑) ∨ (𝜒 → 𝜃))))
5	3, 4	anbi12i 630	. 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)))))
6		an42 657	. 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)))))
7	1, 5, 6	3bitri 300	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 209   ∧ wa 399   ∨ wo 847  if-wif 1063

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 210  df-an 400  df-or 848  df-ifp 1064

This theorem is referenced by: (None)