Theorem ifpororb 39869

Description: Factor conditional logic operator over disjunction in terms 2 and 3. (Contributed by RP, 21-Apr-2020.)

Assertion

Ref	Expression
ifpororb	⊢ (if-(𝜑, (𝜓 ∨ 𝜒), (𝜃 ∨ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))

Proof of Theorem ifpororb

Step	Hyp	Ref	Expression
1		dfifp2 1059	. 2 ⊢ (if-(𝜑, (𝜓 ∨ 𝜒), (𝜃 ∨ 𝜏)) ↔ ((𝜑 → (𝜓 ∨ 𝜒)) ∧ (¬ 𝜑 → (𝜃 ∨ 𝜏))))
2		df-or 844	. . . 4 ⊢ ((𝜓 ∨ 𝜒) ↔ (¬ 𝜓 → 𝜒))
3	2	imbi2i 338	. . 3 ⊢ ((𝜑 → (𝜓 ∨ 𝜒)) ↔ (𝜑 → (¬ 𝜓 → 𝜒)))
4		df-or 844	. . . 4 ⊢ ((𝜃 ∨ 𝜏) ↔ (¬ 𝜃 → 𝜏))
5	4	imbi2i 338	. . 3 ⊢ ((¬ 𝜑 → (𝜃 ∨ 𝜏)) ↔ (¬ 𝜑 → (¬ 𝜃 → 𝜏)))
6	3, 5	anbi12i 628	. 2 ⊢ (((𝜑 → (𝜓 ∨ 𝜒)) ∧ (¬ 𝜑 → (𝜃 ∨ 𝜏))) ↔ ((𝜑 → (¬ 𝜓 → 𝜒)) ∧ (¬ 𝜑 → (¬ 𝜃 → 𝜏))))
7		ifpimimb 39868	. . 3 ⊢ (if-(𝜑, (¬ 𝜓 → 𝜒), (¬ 𝜃 → 𝜏)) ↔ (if-(𝜑, ¬ 𝜓, ¬ 𝜃) → if-(𝜑, 𝜒, 𝜏)))
8		dfifp2 1059	. . 3 ⊢ (if-(𝜑, (¬ 𝜓 → 𝜒), (¬ 𝜃 → 𝜏)) ↔ ((𝜑 → (¬ 𝜓 → 𝜒)) ∧ (¬ 𝜑 → (¬ 𝜃 → 𝜏))))
9		imor 849	. . . 4 ⊢ ((if-(𝜑, ¬ 𝜓, ¬ 𝜃) → if-(𝜑, 𝜒, 𝜏)) ↔ (¬ if-(𝜑, ¬ 𝜓, ¬ 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
10		ifpnot23d 39849	. . . . 5 ⊢ (¬ if-(𝜑, ¬ 𝜓, ¬ 𝜃) ↔ if-(𝜑, 𝜓, 𝜃))
11	10	orbi1i 910	. . . 4 ⊢ ((¬ if-(𝜑, ¬ 𝜓, ¬ 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
12	9, 11	bitri 277	. . 3 ⊢ ((if-(𝜑, ¬ 𝜓, ¬ 𝜃) → if-(𝜑, 𝜒, 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
13	7, 8, 12	3bitr3i 303	. 2 ⊢ (((𝜑 → (¬ 𝜓 → 𝜒)) ∧ (¬ 𝜑 → (¬ 𝜃 → 𝜏))) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
14	1, 6, 13	3bitri 299	1 ⊢ (if-(𝜑, (𝜓 ∨ 𝜒), (𝜃 ∨ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398   ∨ wo 843  if-wif 1057

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 209  df-an 399  df-or 844  df-ifp 1058

This theorem is referenced by:  ifpananb  39870