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Theorem ifp1bi 40253
 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)) ↔ ((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜑𝜓) ∨ (𝜃𝜒))) ∧ (((𝜓𝜑) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒)))))

Proof of Theorem ifp1bi
StepHypRef 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 40252 . . . 4 ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜓𝜑) ∨ (𝜃𝜒)) ∧ ((𝜑𝜓) ∨ (𝜒𝜃))))
32biancomi 466 . . 3 ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒))))
4 ifpim1g 40252 . . 3 ((if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃)) ↔ (((𝜑𝜓) ∨ (𝜃𝜒)) ∧ ((𝜓𝜑) ∨ (𝜒𝜃))))
53, 4anbi12i 629 . 2 (((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ∧ (if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃))) ↔ ((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒))) ∧ (((𝜑𝜓) ∨ (𝜃𝜒)) ∧ ((𝜓𝜑) ∨ (𝜒𝜃)))))
6 an42 656 . 2 (((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒))) ∧ (((𝜑𝜓) ∨ (𝜃𝜒)) ∧ ((𝜓𝜑) ∨ (𝜒𝜃)))) ↔ ((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜑𝜓) ∨ (𝜃𝜒))) ∧ (((𝜓𝜑) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒)))))
71, 5, 63bitri 300 1 ((if-(𝜑, 𝜒, 𝜃) ↔ if-(𝜓, 𝜒, 𝜃)) ↔ ((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜑𝜓) ∨ (𝜃𝜒))) ∧ (((𝜓𝜑) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒)))))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   ∨ wo 844  if-wif 1058 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 845  df-ifp 1059 This theorem is referenced by: (None)
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