Theorem cases2 1042

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 1021	. 2 ⊢ (((𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))) ∧ (¬ 𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))) ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))
2		dedlema 1040	. . . 4 ⊢ (𝜑 → (𝜓 ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
3	2	pm5.74i 273	. . 3 ⊢ ((𝜑 → 𝜓) ↔ (𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
4		dedlemb 1041	. . . 4 ⊢ (¬ 𝜑 → (𝜒 ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
5	4	pm5.74i 273	. . 3 ⊢ ((¬ 𝜑 → 𝜒) ↔ (¬ 𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))))
6	3, 5	anbi12i 628	. 2 ⊢ (((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)) ↔ ((𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑))) ∧ (¬ 𝜑 → ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))))
7		ancom 463	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
8		ancom 463	. . 3 ⊢ ((¬ 𝜑 ∧ 𝜒) ↔ (𝜒 ∧ ¬ 𝜑))
9	7, 8	orbi12i 911	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) ↔ ((𝜓 ∧ 𝜑) ∨ (𝜒 ∧ ¬ 𝜑)))
10	1, 6, 9	3bitr4ri 306	1 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) ↔ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))

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

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

This theorem is referenced by:  dfbi3  1044  dfifp2  1059  ifval  4507  ifpidg  39855  ifpim123g  39864