Theorem sbied

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie ) Usage of this theorem is discouraged because it depends on ax-13 . See sbiedw , sbiedvw for variants using disjoint variables, but requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Mario Carneiro, 4-Oct-2016) (Proof shortened by Wolf Lammen, 24-Jun-2018) (New usage is discouraged.)

	Ref	Expression
Hypotheses	sbied.1	|- F/ x ph
	sbied.2	|- ( ph -> F/ x ch )
	sbied.3	|- ( ph -> ( x = y -> ( ps <-> ch ) ) )
Assertion	sbied	|- ( ph -> ( [ y / x ] ps <-> ch ) )

Proof

Step	Hyp	Ref	Expression
1		sbied.1	|- F/ x ph
2		sbied.2	|- ( ph -> F/ x ch )
3		sbied.3	|- ( ph -> ( x = y -> ( ps <-> ch ) ) )
4	1	sbrim	|- ( [ y / x ] ( ph -> ps ) <-> ( ph -> [ y / x ] ps ) )
5	1 2	nfim1	|- F/ x ( ph -> ch )
6	3	com12	|- ( x = y -> ( ph -> ( ps <-> ch ) ) )
7	6	pm5.74d	|- ( x = y -> ( ( ph -> ps ) <-> ( ph -> ch ) ) )
8	5 7	sbie	|- ( [ y / x ] ( ph -> ps ) <-> ( ph -> ch ) )
9	4 8	bitr3i	|- ( ( ph -> [ y / x ] ps ) <-> ( ph -> ch ) )
10	9	pm5.74ri	|- ( ph -> ( [ y / x ] ps <-> ch ) )