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) Revise df-sb again. (Revised by Wolf Lammen, 4-Jun-2026)

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

Proof

Step	Hyp	Ref	Expression
1		sbtlem.1	|- ph
2	1	sbtlem	|- A. y ( y = t -> A. x ( x = y -> ph ) )
3	1	sbtlem	|- A. z ( z = t -> A. x ( x = z -> ph ) )
4	2 3	pm3.2i	|- ( A. y ( y = t -> A. x ( x = y -> ph ) ) /\ A. z ( z = t -> A. x ( x = z -> ph ) ) )
5		df-sb	|- ( [ t / x ] ph <-> ( A. y ( y = t -> A. x ( x = y -> ph ) ) /\ A. z ( z = t -> A. x ( x = z -> ph ) ) ) )
6	4 5	mpbir	|- [ t / x ] ph