Theorem sbequ6

Substitution does not change a distinctor. Usage of this theorem is discouraged because it depends on [ax-13]. (Contributed by NM, 5-Aug-1993.) (New usage is discouraged.)

Conclusion

\( \vdash [ w' / z ] \, (\neg \forall x \, x = y) \leftrightarrow \neg \forall x \, x = y \)

Proof

nfnae

sbf _