Theorem ifpid2g 40998

Description: Restate wff as conditional logic operator. (Contributed by RP, 20-Apr-2020.)

Assertion

Ref	Expression
ifpid2g	⊢ ((𝜓 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((𝜓 → (𝜑 ∨ 𝜒)) ∧ (𝜒 → (𝜑 ∨ 𝜓))))

Proof of Theorem ifpid2g

Step	Hyp	Ref	Expression
1		ifpidg 40996	. 2 ⊢ ((𝜓 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((((𝜑 ∧ 𝜓) → 𝜓) ∧ ((𝜑 ∧ 𝜓) → 𝜓)) ∧ ((𝜒 → (𝜑 ∨ 𝜓)) ∧ (𝜓 → (𝜑 ∨ 𝜒)))))
2		simpr 484	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜓)
3	2, 2	pm3.2i 470	. . 3 ⊢ (((𝜑 ∧ 𝜓) → 𝜓) ∧ ((𝜑 ∧ 𝜓) → 𝜓))
4	3	biantrur 530	. 2 ⊢ (((𝜒 → (𝜑 ∨ 𝜓)) ∧ (𝜓 → (𝜑 ∨ 𝜒))) ↔ ((((𝜑 ∧ 𝜓) → 𝜓) ∧ ((𝜑 ∧ 𝜓) → 𝜓)) ∧ ((𝜒 → (𝜑 ∨ 𝜓)) ∧ (𝜓 → (𝜑 ∨ 𝜒)))))
5		ancom 460	. 2 ⊢ (((𝜒 → (𝜑 ∨ 𝜓)) ∧ (𝜓 → (𝜑 ∨ 𝜒))) ↔ ((𝜓 → (𝜑 ∨ 𝜒)) ∧ (𝜒 → (𝜑 ∨ 𝜓))))
6	1, 4, 5	3bitr2i 298	1 ⊢ ((𝜓 ↔ if-(𝜑, 𝜓, 𝜒)) ↔ ((𝜓 → (𝜑 ∨ 𝜒)) ∧ (𝜒 → (𝜑 ∨ 𝜓))))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843  if-wif 1059

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 206  df-an 396  df-or 844  df-ifp 1060

This theorem is referenced by: (None)