Metamath Proof Explorer


Theorem ecase23d

Description: Deduction for elimination by cases. (Contributed by NM, 22-Apr-1994)

Ref Expression
Hypotheses ecase23d.1 ⊢ ( 𝜑 → ¬ 𝜒 )
ecase23d.2 ⊢ ( 𝜑 → ¬ 𝜃 )
ecase23d.3 ⊢ ( 𝜑 → ( 𝜓 ∨ 𝜒 ∨ 𝜃 ) )
Assertion ecase23d ( 𝜑 → 𝜓 )

Proof

Step Hyp Ref Expression
1 ecase23d.1 ⊢ ( 𝜑 → ¬ 𝜒 )
2 ecase23d.2 ⊢ ( 𝜑 → ¬ 𝜃 )
3 ecase23d.3 ⊢ ( 𝜑 → ( 𝜓 ∨ 𝜒 ∨ 𝜃 ) )
4 3orass ⊢ ( ( 𝜓 ∨ 𝜒 ∨ 𝜃 ) ↔ ( 𝜓 ∨ ( 𝜒 ∨ 𝜃 ) ) )
5 3 4 sylib ⊢ ( 𝜑 → ( 𝜓 ∨ ( 𝜒 ∨ 𝜃 ) ) )
6 ioran ⊢ ( ¬ ( 𝜒 ∨ 𝜃 ) ↔ ( ¬ 𝜒 ∧ ¬ 𝜃 ) )
7 1 2 6 sylanbrc ⊢ ( 𝜑 → ¬ ( 𝜒 ∨ 𝜃 ) )
8 5 7 olcnd ⊢ ( 𝜑 → 𝜓 )