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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 )