Metamath Proof Explorer


Theorem sbt

Description: A substitution into a theorem yields a theorem. See sbtALT for a shorter proof requiring more axioms. See chvar and chvarv for versions using implicit substitution. (Contributed by NM, 21-Jan-2004) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 20-Jul-2018) Revise df-sb . (Revised by Steven Nguyen, 6-Jul-2023)

Ref Expression
Hypothesis sbt.1
|- ph
Assertion sbt
|- [ t / x ] ph

Proof

Step Hyp Ref Expression
1 sbt.1
 |-  ph
2 1 a1i
 |-  ( x = y -> ph )
3 2 ax-gen
 |-  A. x ( x = y -> ph )
4 3 a1i
 |-  ( y = t -> A. x ( x = y -> ph ) )
5 4 ax-gen
 |-  A. y ( y = t -> A. x ( x = y -> ph ) )
6 df-sb
 |-  ( [ t / x ] ph <-> A. y ( y = t -> A. x ( x = y -> ph ) ) )
7 5 6 mpbir
 |-  [ t / x ] ph