Metamath Proof Explorer


Theorem cases2

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

Ref Expression
Assertion cases2 ( ( ( 𝜑𝜓 ) ∨ ( ¬ 𝜑𝜒 ) ) ↔ ( ( 𝜑𝜓 ) ∧ ( ¬ 𝜑𝜒 ) ) )

Proof

Step Hyp Ref Expression
1 pm4.83 ( ( ( 𝜑 → ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) ∧ ( ¬ 𝜑 → ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) ) ↔ ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) )
2 dedlema ( 𝜑 → ( 𝜓 ↔ ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) )
3 2 pm5.74i ( ( 𝜑𝜓 ) ↔ ( 𝜑 → ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) )
4 dedlemb ( ¬ 𝜑 → ( 𝜒 ↔ ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) )
5 4 pm5.74i ( ( ¬ 𝜑𝜒 ) ↔ ( ¬ 𝜑 → ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) )
6 3 5 anbi12i ( ( ( 𝜑𝜓 ) ∧ ( ¬ 𝜑𝜒 ) ) ↔ ( ( 𝜑 → ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) ∧ ( ¬ 𝜑 → ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) ) ) )
7 ancom ( ( 𝜑𝜓 ) ↔ ( 𝜓𝜑 ) )
8 ancom ( ( ¬ 𝜑𝜒 ) ↔ ( 𝜒 ∧ ¬ 𝜑 ) )
9 7 8 orbi12i ( ( ( 𝜑𝜓 ) ∨ ( ¬ 𝜑𝜒 ) ) ↔ ( ( 𝜓𝜑 ) ∨ ( 𝜒 ∧ ¬ 𝜑 ) ) )
10 1 6 9 3bitr4ri ( ( ( 𝜑𝜓 ) ∨ ( ¬ 𝜑𝜒 ) ) ↔ ( ( 𝜑𝜓 ) ∧ ( ¬ 𝜑𝜒 ) ) )