Metamath Proof Explorer


Theorem pm2.61dane

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 30-Nov-2011)

Ref Expression
Hypotheses pm2.61dane.1 ( ( 𝜑𝐴 = 𝐵 ) → 𝜓 )
pm2.61dane.2 ( ( 𝜑𝐴𝐵 ) → 𝜓 )
Assertion pm2.61dane ( 𝜑𝜓 )

Proof

Step Hyp Ref Expression
1 pm2.61dane.1 ( ( 𝜑𝐴 = 𝐵 ) → 𝜓 )
2 pm2.61dane.2 ( ( 𝜑𝐴𝐵 ) → 𝜓 )
3 1 ex ( 𝜑 → ( 𝐴 = 𝐵𝜓 ) )
4 2 ex ( 𝜑 → ( 𝐴𝐵𝜓 ) )
5 3 4 pm2.61dne ( 𝜑𝜓 )