Metamath Proof Explorer

Theorem sb2

Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition ( sb6 ) or a non-freeness hypothesis ( sb6f ). Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 13-May-1993) Revise df-sb . (Revised by Wolf Lammen, 26-Jul-2023) (New usage is discouraged.)

Ref Expression
Assertion sb2 ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({x}={y}\to {\phi }\right)\to \left[{y}/{x}\right]{\phi }$

Proof

Step Hyp Ref Expression
1 pm2.27 ${⊢}{x}={y}\to \left(\left({x}={y}\to {\phi }\right)\to {\phi }\right)$
2 1 al2imi ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}{x}={y}\to \left(\forall {x}\phantom{\rule{.4em}{0ex}}\left({x}={y}\to {\phi }\right)\to \forall {x}\phantom{\rule{.4em}{0ex}}{\phi }\right)$
3 stdpc4 ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}{\phi }\to \left[{y}/{x}\right]{\phi }$
4 2 3 syl6 ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}{x}={y}\to \left(\forall {x}\phantom{\rule{.4em}{0ex}}\left({x}={y}\to {\phi }\right)\to \left[{y}/{x}\right]{\phi }\right)$
5 sb4b ${⊢}¬\forall {x}\phantom{\rule{.4em}{0ex}}{x}={y}\to \left(\left[{y}/{x}\right]{\phi }↔\forall {x}\phantom{\rule{.4em}{0ex}}\left({x}={y}\to {\phi }\right)\right)$
6 5 biimprd ${⊢}¬\forall {x}\phantom{\rule{.4em}{0ex}}{x}={y}\to \left(\forall {x}\phantom{\rule{.4em}{0ex}}\left({x}={y}\to {\phi }\right)\to \left[{y}/{x}\right]{\phi }\right)$
7 4 6 pm2.61i ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({x}={y}\to {\phi }\right)\to \left[{y}/{x}\right]{\phi }$