Metamath Proof Explorer


Theorem nf5d

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016)

Ref Expression
Hypotheses nf5d.1 xφ
nf5d.2 φψxψ
Assertion nf5d φxψ

Proof

Step Hyp Ref Expression
1 nf5d.1 xφ
2 nf5d.2 φψxψ
3 1 2 alrimi φxψxψ
4 nf5-1 xψxψxψ
5 3 4 syl φxψ