Theorem wl-ax11-lem10 33371

Description: We now have prepared everything. The unwanted variable u is just in one place left. pm2.61 183 can be used in conjunction with wl-ax11-lem9 33370 to eliminate the second antecedent. Missing is something along the lines of ax-6 1888, so we could remove the first antecedent. But the Metamath axioms cannot accomplish this. Such a rule must reside one abstraction level higher than all others: It says that a distinctor implies a distinct variable condition on its contained setvar. This is only needed if such conditions are required, as ax-11v does. The result of this study is for me, that you cannot introduce a setvar capturing this condition, and hope to eliminate it later. (Contributed by Wolf Lammen, 30-Jun-2019.)

Assertion

Ref	Expression
wl-ax11-lem10	|- ( A. y y = u -> ( -. A. x x = y -> ( A. y A. x ph -> A. x A. y ph ) ) )

Distinct variable group:   x, u

Allowed substitution hints:   ph( x, y, u)

Proof of Theorem wl-ax11-lem10

Step	Hyp	Ref	Expression
1		wl-ax11-lem8 33369	. . . . 5 |- ( ( A. u u = y /\ -. A. x x = y ) -> ( A. u A. x [ u / y ] ph <-> A. y A. x ph ) )
2		wl-ax11-lem6 33367	. . . . 5 |- ( ( A. u u = y /\ -. A. x x = y ) -> ( A. u A. x [ u / y ] ph <-> A. x A. y ph ) )
3	1, 2	bitr3d 270	. . . 4 |- ( ( A. u u = y /\ -. A. x x = y ) -> ( A. y A. x ph <-> A. x A. y ph ) )
4	3	biimpd 219	. . 3 |- ( ( A. u u = y /\ -. A. x x = y ) -> ( A. y A. x ph -> A. x A. y ph ) )
5	4	ex 450	. 2 |- ( A. u u = y -> ( -. A. x x = y -> ( A. y A. x ph -> A. x A. y ph ) ) )
6	5	aecoms 2312	1 |- ( A. y y = u -> ( -. A. x x = y -> ( A. y A. x ph -> A. x A. y ph ) ) )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 384   A.wal 1481   [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246  ax-wl-11v 33361

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)