Metamath Proof Explorer


Theorem axc14

Description: Axiom ax-c14 is redundant if we assume ax-5 . Remark 9.6 in Megill p. 448 (p. 16 of the preprint), regarding axiom scheme C14'.

Note that w is a dummy variable introduced in the proof. Its purpose is to satisfy the distinct variable requirements of dveel2 and ax-5 . By the end of the proof it has vanished, and the final theorem has no distinct variable requirements. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 29-Jun-1995) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion axc14
|- ( -. A. z z = x -> ( -. A. z z = y -> ( x e. y -> A. z x e. y ) ) )

Proof

Step Hyp Ref Expression
1 hbn1
 |-  ( -. A. z z = y -> A. z -. A. z z = y )
2 dveel2
 |-  ( -. A. z z = y -> ( w e. y -> A. z w e. y ) )
3 1 2 hbim1
 |-  ( ( -. A. z z = y -> w e. y ) -> A. z ( -. A. z z = y -> w e. y ) )
4 elequ1
 |-  ( w = x -> ( w e. y <-> x e. y ) )
5 4 imbi2d
 |-  ( w = x -> ( ( -. A. z z = y -> w e. y ) <-> ( -. A. z z = y -> x e. y ) ) )
6 3 5 dvelim
 |-  ( -. A. z z = x -> ( ( -. A. z z = y -> x e. y ) -> A. z ( -. A. z z = y -> x e. y ) ) )
7 nfa1
 |-  F/ z A. z z = y
8 7 nfn
 |-  F/ z -. A. z z = y
9 8 19.21
 |-  ( A. z ( -. A. z z = y -> x e. y ) <-> ( -. A. z z = y -> A. z x e. y ) )
10 6 9 syl6ib
 |-  ( -. A. z z = x -> ( ( -. A. z z = y -> x e. y ) -> ( -. A. z z = y -> A. z x e. y ) ) )
11 10 pm2.86d
 |-  ( -. A. z z = x -> ( -. A. z z = y -> ( x e. y -> A. z x e. y ) ) )