Metamath Proof Explorer


Theorem hbim

Description: If x is not free in ph and ps , it is not free in ( ph -> ps ) . (Contributed by NM, 24-Jan-1993) (Proof shortened by Mel L. O'Cat, 3-Mar-2008) (Proof shortened by Wolf Lammen, 1-Jan-2018)

Ref Expression
Hypotheses hbim.1 ( 𝜑 → ∀ 𝑥 𝜑 )
hbim.2 ( 𝜓 → ∀ 𝑥 𝜓 )
Assertion hbim ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 hbim.1 ( 𝜑 → ∀ 𝑥 𝜑 )
2 hbim.2 ( 𝜓 → ∀ 𝑥 𝜓 )
3 2 a1i ( 𝜑 → ( 𝜓 → ∀ 𝑥 𝜓 ) )
4 1 3 hbim1 ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) )