Metamath Proof Explorer


Theorem wl-ax11-lem10

Description: We now have prepared everything. The unwanted variable u is just in one place left. pm2.61 can be used in conjunction with wl-ax11-lem9 to eliminate the second antecedent. Missing is something along the lines of ax-6 , 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)

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

Proof

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