Theorem cases2 1005

Description: Case disjunction according to the value of 𝜑. (Contributed by BJ, 6-Apr-2019.) (Proof shortened by Wolf Lammen, 2-Jan-2020.)

Assertion

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

Proof of Theorem cases2

Step	Hyp	Ref	Expression
1		pm3.4 582	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → (𝜑 → 𝜓))
2		pm2.24 120	. . . . 5 ⊢ (𝜑 → (¬ 𝜑 → 𝜒))
3	2	adantr 480	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → (¬ 𝜑 → 𝜒))
4	1, 3	jca 553	. . 3 ⊢ ((𝜑 ∧ 𝜓) → ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
5		pm2.21 119	. . . . 5 ⊢ (¬ 𝜑 → (𝜑 → 𝜓))
6	5	adantr 480	. . . 4 ⊢ ((¬ 𝜑 ∧ 𝜒) → (𝜑 → 𝜓))
7		pm3.4 582	. . . 4 ⊢ ((¬ 𝜑 ∧ 𝜒) → (¬ 𝜑 → 𝜒))
8	6, 7	jca 553	. . 3 ⊢ ((¬ 𝜑 ∧ 𝜒) → ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
9	4, 8	jaoi 393	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) → ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
10		pm2.27 41	. . . . . 6 ⊢ (𝜑 → ((𝜑 → 𝜓) → 𝜓))
11	10	imdistani 722	. . . . 5 ⊢ ((𝜑 ∧ (𝜑 → 𝜓)) → (𝜑 ∧ 𝜓))
12	11	orcd 406	. . . 4 ⊢ ((𝜑 ∧ (𝜑 → 𝜓)) → ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)))
13	12	adantrr 749	. . 3 ⊢ ((𝜑 ∧ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒))) → ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)))
14		pm2.27 41	. . . . . 6 ⊢ (¬ 𝜑 → ((¬ 𝜑 → 𝜒) → 𝜒))
15	14	imdistani 722	. . . . 5 ⊢ ((¬ 𝜑 ∧ (¬ 𝜑 → 𝜒)) → (¬ 𝜑 ∧ 𝜒))
16	15	olcd 407	. . . 4 ⊢ ((¬ 𝜑 ∧ (¬ 𝜑 → 𝜒)) → ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)))
17	16	adantrl 748	. . 3 ⊢ ((¬ 𝜑 ∧ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒))) → ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)))
18	13, 17	pm2.61ian 827	. 2 ⊢ (((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)) → ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)))
19	9, 18	impbii 198	1 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) ↔ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∨ wo 382   ∧ wa 383

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 196  df-or 384  df-an 385

This theorem is referenced by:  dfifp2  1008  ifval  4077  ifpidg  36855  ifpim123g  36864