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Metamath Proof Explorer

Theorem ecase3ad

Description: Deduction for elimination by cases. (Contributed by NM, 24-May-2013) (Proof shortened by Wolf Lammen, 20-Sep-2024)

Ref Expression
Hypotheses ecase3ad.1 ⊢ φ → ψ → θ
ecase3ad.2 ⊢ φ → χ → θ
ecase3ad.3 ⊢ φ → ¬ ψ ∧ ¬ χ → θ
Assertion ecase3ad ⊢ φ → θ

Proof

Step Hyp Ref Expression
1 ecase3ad.1 ⊢ φ → ψ → θ
2 ecase3ad.2 ⊢ φ → χ → θ
3 ecase3ad.3 ⊢ φ → ¬ ψ ∧ ¬ χ → θ
4 1 imp ⊢ φ ∧ ψ → θ
5 2 imp ⊢ φ ∧ χ → θ
6 3 imp ⊢ φ ∧ ¬ ψ ∧ ¬ χ → θ
7 4 5 6 pm2.61ddan ⊢ φ → θ