Metamath Proof Explorer


Theorem nf3or

Description: If x is not free in ph , ps , and ch , then it is not free in ( ph \/ ps \/ ch ) . (Contributed by Mario Carneiro, 11-Aug-2016)

Ref Expression
Hypotheses nf.1
|- F/ x ph
nf.2
|- F/ x ps
nf.3
|- F/ x ch
Assertion nf3or
|- F/ x ( ph \/ ps \/ ch )

Proof

Step Hyp Ref Expression
1 nf.1
 |-  F/ x ph
2 nf.2
 |-  F/ x ps
3 nf.3
 |-  F/ x ch
4 df-3or
 |-  ( ( ph \/ ps \/ ch ) <-> ( ( ph \/ ps ) \/ ch ) )
5 1 2 nfor
 |-  F/ x ( ph \/ ps )
6 5 3 nfor
 |-  F/ x ( ( ph \/ ps ) \/ ch )
7 4 6 nfxfr
 |-  F/ x ( ph \/ ps \/ ch )