Theorem rgenm 3525

Description: Generalization rule that eliminates an inhabited class requirement. (Contributed by Jim Kingdon, 5-Aug-2018.)

Hypothesis

Ref	Expression
rgenm.1	|- ( ( E. x x e. A /\ x e. A ) -> ph )

Assertion

Ref	Expression
rgenm	|- A. x e. A ph

Distinct variable group:   x, A

Allowed substitution hint:   ph( x)

Proof of Theorem rgenm

Step	Hyp	Ref	Expression
1		nfe1 1496	. . . . 5 |- F/ x E. x x e. A
2		rgenm.1	. . . . . 6 |- ( ( E. x x e. A /\ x e. A ) -> ph )
3	2	ex 115	. . . . 5 |- ( E. x x e. A -> ( x e. A -> ph ) )
4	1, 3	alrimi 1522	. . . 4 |- ( E. x x e. A -> A. x ( x e. A -> ph ) )
5		19.38 1676	. . . 4 |- ( ( E. x x e. A -> A. x ( x e. A -> ph ) ) -> A. x ( x e. A -> ( x e. A -> ph ) ) )
6	4, 5	ax-mp 5	. . 3 |- A. x ( x e. A -> ( x e. A -> ph ) )
7		pm5.4 249	. . . 4 |- ( ( x e. A -> ( x e. A -> ph ) ) <-> ( x e. A -> ph ) )
8	7	albii 1470	. . 3 |- ( A. x ( x e. A -> ( x e. A -> ph ) ) <-> A. x ( x e. A -> ph ) )
9	6, 8	mpbi 145	. 2 |- A. x ( x e. A -> ph )
10		df-ral 2460	. 2 |- ( A. x e. A ph <-> A. x ( x e. A -> ph ) )
11	9, 10	mpbir 146	1 |- A. x e. A ph

Syntax hints:   -> wi 4   /\ wa 104   A.wal 1351   E.wex 1492   e. wcel 2148   A.wral 2455

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 1447  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-4 1510  ax-ial 1534  ax-i5r 1535

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-ral 2460

This theorem is referenced by: (None)