Metamath Proof Explorer


Theorem ecase2d

Description: Deduction for elimination by cases. (Contributed by NM, 21-Apr-1994) (Proof shortened by Wolf Lammen, 19-Sep-2024)

Ref Expression
Hypotheses ecase2d.1 ( 𝜑𝜓 )
ecase2d.2 ( 𝜑 → ¬ ( 𝜓𝜒 ) )
ecase2d.3 ( 𝜑 → ¬ ( 𝜓𝜃 ) )
ecase2d.4 ( 𝜑 → ( 𝜏 ∨ ( 𝜒𝜃 ) ) )
Assertion ecase2d ( 𝜑𝜏 )

Proof

Step Hyp Ref Expression
1 ecase2d.1 ( 𝜑𝜓 )
2 ecase2d.2 ( 𝜑 → ¬ ( 𝜓𝜒 ) )
3 ecase2d.3 ( 𝜑 → ¬ ( 𝜓𝜃 ) )
4 ecase2d.4 ( 𝜑 → ( 𝜏 ∨ ( 𝜒𝜃 ) ) )
5 1 2 mpnanrd ( 𝜑 → ¬ 𝜒 )
6 1 3 mpnanrd ( 𝜑 → ¬ 𝜃 )
7 4 ord ( 𝜑 → ( ¬ 𝜏 → ( 𝜒𝜃 ) ) )
8 5 6 7 mtord ( 𝜑 → ¬ ¬ 𝜏 )
9 8 notnotrd ( 𝜑𝜏 )