Metamath Proof Explorer


Theorem 4cases

Description: Inference eliminating two antecedents from the four possible cases that result from their true/false combinations. (Contributed by NM, 25-Oct-2003)

Ref Expression
Hypotheses 4cases.1 ( ( 𝜑𝜓 ) → 𝜒 )
4cases.2 ( ( 𝜑 ∧ ¬ 𝜓 ) → 𝜒 )
4cases.3 ( ( ¬ 𝜑𝜓 ) → 𝜒 )
4cases.4 ( ( ¬ 𝜑 ∧ ¬ 𝜓 ) → 𝜒 )
Assertion 4cases 𝜒

Proof

Step Hyp Ref Expression
1 4cases.1 ( ( 𝜑𝜓 ) → 𝜒 )
2 4cases.2 ( ( 𝜑 ∧ ¬ 𝜓 ) → 𝜒 )
3 4cases.3 ( ( ¬ 𝜑𝜓 ) → 𝜒 )
4 4cases.4 ( ( ¬ 𝜑 ∧ ¬ 𝜓 ) → 𝜒 )
5 1 3 pm2.61ian ( 𝜓𝜒 )
6 2 4 pm2.61ian ( ¬ 𝜓𝜒 )
7 5 6 pm2.61i 𝜒