Theorem cases 1040

Description: Case disjunction according to the value of 𝜑. (Contributed by NM, 25-Apr-2019.)

Hypotheses

Ref	Expression
cases.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))
cases.2	⊢ (¬ 𝜑 → (𝜓 ↔ 𝜃))

Assertion

Ref	Expression
cases	⊢ (𝜓 ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))

Proof of Theorem cases

Step	Hyp	Ref	Expression
1		exmid 892	. . 3 ⊢ (𝜑 ∨ ¬ 𝜑)
2	1	biantrur 531	. 2 ⊢ (𝜓 ↔ ((𝜑 ∨ ¬ 𝜑) ∧ 𝜓))
3		andir 1006	. 2 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ↔ ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))
4		cases.1	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
5	4	pm5.32i 575	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒))
6		cases.2	. . . 4 ⊢ (¬ 𝜑 → (𝜓 ↔ 𝜃))
7	6	pm5.32i 575	. . 3 ⊢ ((¬ 𝜑 ∧ 𝜓) ↔ (¬ 𝜑 ∧ 𝜃))
8	5, 7	orbi12i 912	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)) ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))
9	2, 3, 8	3bitri 297	1 ⊢ (𝜓 ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396   ∨ wo 844

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 397  df-or 845

This theorem is referenced by:  casesifp  1076  elimif  4496  elim2if  30887  wl-ifpimpr  35637