Theorem dfsb1

Description: Alternate definition of substitution. Remark 9.1 in Megill p. 447 (p. 15 of the preprint). This was the original definition before df-sb . Note that it does not require dummy variables in its definiens; this is done by having x free in the first conjunct and bound in the second. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by BJ, 9-Jul-2023) Revise df-sb . (Revised by Wolf Lammen, 29-Jul-2023) (New usage is discouraged.)

	Ref	Expression
Assertion	dfsb1	|- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )

Proof

Step	Hyp	Ref	Expression
1		sbequ2	|- ( x = y -> ( [ y / x ] ph -> ph ) )
2	1	com12	|- ( [ y / x ] ph -> ( x = y -> ph ) )
3		sb1	|- ( [ y / x ] ph -> E. x ( x = y /\ ph ) )
4	2 3	jca	|- ( [ y / x ] ph -> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
5		id	|- ( x = y -> x = y )
6		sbequ1	|- ( x = y -> ( ph -> [ y / x ] ph ) )
7	5 6	embantd	|- ( x = y -> ( ( x = y -> ph ) -> [ y / x ] ph ) )
8	7	sps	|- ( A. x x = y -> ( ( x = y -> ph ) -> [ y / x ] ph ) )
9	8	adantrd	|- ( A. x x = y -> ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> [ y / x ] ph ) )
10		sb3	|- ( -. A. x x = y -> ( E. x ( x = y /\ ph ) -> [ y / x ] ph ) )
11	10	adantld	|- ( -. A. x x = y -> ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> [ y / x ] ph ) )
12	9 11	pm2.61i	|- ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) -> [ y / x ] ph )
13	4 12	impbii	|- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )