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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 )