Theorem sbbidv 1899

Description: Deduction substituting both sides of a biconditional, with ph and x disjoint. See also sbbid 1860. (Contributed by Wolf Lammen, 6-May-2023.) (Proof shortened by Steven Nguyen, 6-Jul-2023.)

Hypothesis

Ref	Expression
sbbidv.1	|- ( ph -> ( ps <-> ch ) )

Assertion

Ref	Expression
sbbidv	|- ( ph -> ( [ t / x ] ps <-> [ t / x ] ch ) )

Distinct variable group:   ph, x

Allowed substitution hints:   ph( t)   ps( x, t)   ch( x, t)

Proof of Theorem sbbidv

Step	Hyp	Ref	Expression
1		sbbidv.1	. . 3 |- ( ph -> ( ps <-> ch ) )
2	1	alrimiv 1888	. 2 |- ( ph -> A. x ( ps <-> ch ) )
3		spsbbi 1858	. 2 |- ( A. x ( ps <-> ch ) -> ( [ t / x ] ps <-> [ t / x ] ch ) )
4	2, 3	syl 14	1 |- ( ph -> ( [ t / x ] ps <-> [ t / x ] ch ) )

Syntax hints:   -> wi 4   <-> wb 105   A.wal 1362   [wsb 1776

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-17 1540  ax-ial 1548

This theorem depends on definitions:  df-bi 117  df-sb 1777

This theorem is referenced by:  eqabdv  2325