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Theorem cases 1004
 Description: Case disjunction according to the value of 𝜑. (Contributed by NM, 25-Apr-2019.)
Hypotheses
Ref Expression
cases.1 (𝜑 → (𝜓𝜒))
cases.2 𝜑 → (𝜓𝜃))
Assertion
Ref Expression
cases (𝜓 ↔ ((𝜑𝜒) ∨ (¬ 𝜑𝜃)))

Proof of Theorem cases
StepHypRef Expression
1 exmid 430 . . 3 (𝜑 ∨ ¬ 𝜑)
21biantrur 526 . 2 (𝜓 ↔ ((𝜑 ∨ ¬ 𝜑) ∧ 𝜓))
3 andir 908 . 2 (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ↔ ((𝜑𝜓) ∨ (¬ 𝜑𝜓)))
4 cases.1 . . . 4 (𝜑 → (𝜓𝜒))
54pm5.32i 667 . . 3 ((𝜑𝜓) ↔ (𝜑𝜒))
6 cases.2 . . . 4 𝜑 → (𝜓𝜃))
76pm5.32i 667 . . 3 ((¬ 𝜑𝜓) ↔ (¬ 𝜑𝜃))
85, 7orbi12i 542 . 2 (((𝜑𝜓) ∨ (¬ 𝜑𝜓)) ↔ ((𝜑𝜒) ∨ (¬ 𝜑𝜃)))
92, 3, 83bitri 285 1 (𝜓 ↔ ((𝜑𝜒) ∨ (¬ 𝜑𝜃)))
 Colors of variables: wff setvar class 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:  casesifp  1020  elimif  4072  elim2if  28747
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