Theorem cases 1058

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 908	. . 3 ⊢ (𝜑 ∨ ¬ 𝜑)
2	1	biantrur 540	. 2 ⊢ (𝜓 ↔ ((𝜑 ∨ ¬ 𝜑) ∧ 𝜓))
3		andir 1026	. 2 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ↔ ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))
4		cases.1	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
5	4	pm5.32i 585	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒))
6		cases.2	. . . 4 ⊢ (¬ 𝜑 → (𝜓 ↔ 𝜃))
7	6	pm5.32i 585	. . 3 ⊢ ((¬ 𝜑 ∧ 𝜓) ↔ (¬ 𝜑 ∧ 𝜃))
8	5, 7	orbi12i 928	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)) ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))
9	2, 3, 8	3bitri 300	1 ⊢ (𝜓 ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862

This theorem is used by:  casesifp  1094  elimif  4519  elim2if  33074  wl-ifpimpr  38309