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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 )