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Theorem ifp1bi 38349
 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 663 . 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 38348 . . . 4 ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜓𝜑) ∨ (𝜃𝜒)) ∧ ((𝜑𝜓) ∨ (𝜒𝜃))))
3 ancom 465 . . . 4 ((((𝜓𝜑) ∨ (𝜃𝜒)) ∧ ((𝜑𝜓) ∨ (𝜒𝜃))) ↔ (((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒))))
42, 3bitri 264 . . 3 ((if-(𝜑, 𝜒, 𝜃) → if-(𝜓, 𝜒, 𝜃)) ↔ (((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒))))
5 ifpim1g 38348 . . 3 ((if-(𝜓, 𝜒, 𝜃) → if-(𝜑, 𝜒, 𝜃)) ↔ (((𝜑𝜓) ∨ (𝜃𝜒)) ∧ ((𝜓𝜑) ∨ (𝜒𝜃))))
64, 5anbi12i 735 . 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)))))
7 an42 901 . 2 (((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒))) ∧ (((𝜑𝜓) ∨ (𝜃𝜒)) ∧ ((𝜓𝜑) ∨ (𝜒𝜃)))) ↔ ((((𝜑𝜓) ∨ (𝜒𝜃)) ∧ ((𝜑𝜓) ∨ (𝜃𝜒))) ∧ (((𝜓𝜑) ∨ (𝜒𝜃)) ∧ ((𝜓𝜑) ∨ (𝜃𝜒)))))
81, 6, 73bitri 286 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 196   ∨ wo 382   ∧ wa 383  if-wif 1050 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 197  df-or 384  df-an 385  df-ifp 1051 This theorem is referenced by: (None)
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