Theorem sbal2

Move quantifier in and out of substitution. (Contributed by NM, 2-Jan-2002.) Remove a distinct variable constraint. (Revised by Wolf Lammen, 24-Dec-2022.) (Proof shortened by Wolf Lammen, 23-Sep-2023.) Usage of this theorem is discouraged because it depends on [ax-13]. Use sbal instead. (New usage is discouraged.)

Conclusion
\( \vdash \neg \forall x \, x = y \rightarrow ([ z' / y ] \, \forall x \, \varphi \leftrightarrow \forall x \, [ z' / y ] \, \varphi) \)
Proof
l0
sbal
l1
a1i _