Metamath Proof Explorer


Theorem nfim

Description: If x is not free in ph and ps , then it is not free in ( ph -> ps ) . Inference associated with nfimt . (Contributed by Mario Carneiro, 11-Aug-2016) (Proof shortened by Wolf Lammen, 2-Jan-2018) df-nf changed. (Revised by Wolf Lammen, 17-Sep-2021)

Ref Expression
Hypotheses nfim.1
|- F/ x ph
nfim.2
|- F/ x ps
Assertion nfim
|- F/ x ( ph -> ps )

Proof

Step Hyp Ref Expression
1 nfim.1
 |-  F/ x ph
2 nfim.2
 |-  F/ x ps
3 nfimt
 |-  ( ( F/ x ph /\ F/ x ps ) -> F/ x ( ph -> ps ) )
4 1 2 3 mp2an
 |-  F/ x ( ph -> ps )