Theorem cases2 1063

Description: Case disjunction according to the value of 𝜑. (Contributed by BJ, 6-Apr-2019.) (Proof shortened by Wolf Lammen, 28-Feb-2022.)

Assertion

Ref	Expression
cases2	⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) ↔ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))

Proof of Theorem cases2

Step	Hyp	Ref	Expression
1		pm4.83 1042	. 2 ⊢ (((𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))) ∧ (¬ 𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))) ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))
2		dedlema 1061	. . . 4 ⊢ (𝜑 → (𝜓 ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
3	2	pm5.74i 274	. . 3 ⊢ ((𝜑 → 𝜓) ↔ (𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
4		dedlemb 1062	. . . 4 ⊢ (¬ 𝜑 → (𝜒 ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
5	4	pm5.74i 274	. . 3 ⊢ ((¬ 𝜑 → 𝜒) ↔ (¬ 𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
6	3, 5	anbi12i 640	. 2 ⊢ (((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)) ↔ ((𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))) ∧ (¬ 𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))))
7		ancom 466	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
8		ancom 466	. . 3 ⊢ ((¬ 𝜑 ∧ 𝜒) ↔ (𝜒 ∧ ¬ 𝜑))
9	7, 8	orbi12i 928	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))
10	1, 6, 9	3bitr4ri 307	1 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) ↔ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))

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:  dfbi3  1065  dfifp2  1080  ifval  4525  ifpidg  44450  ifpim123g  44459