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Metamath Proof Explorer

Theorem ecase3d

Description: Deduction for elimination by cases. (Contributed by NM, 2-May-1996) (Proof shortened by Andrew Salmon, 7-May-2011)

Ref Expression
Hypotheses ecase3d.1 ⊢ φ → ψ → θ
ecase3d.2 ⊢ φ → χ → θ
ecase3d.3 ⊢ φ → ¬ ψ ∨ χ → θ
Assertion ecase3d ⊢ φ → θ

Proof

Step Hyp Ref Expression
1 ecase3d.1 ⊢ φ → ψ → θ
2 ecase3d.2 ⊢ φ → χ → θ
3 ecase3d.3 ⊢ φ → ¬ ψ ∨ χ → θ
4 1 2 jaod ⊢ φ → ψ ∨ χ → θ
5 4 3 pm2.61d ⊢ φ → θ