Metamath Proof Explorer


Theorem exneq

Description: Given any set (the " y " in the statement), there exists a set not equal to it.

The same statement without disjoint variable condition is false, since we do not have E. x -. x = x . This theorem is proved directly from set theory axioms (no class definitions) and does not depend on ax-ext , ax-sep , or ax-pow nor auxiliary logical axiom schemes ax-10 to ax-13 . See dtruALT for a shorter proof using more axioms, and dtruALT2 for a proof using ax-pow instead of ax-pr . (Contributed by NM, 7-Nov-2006) Avoid ax-13 . (Revised by BJ, 31-May-2019) Avoid ax-8 . (Revised by SN, 21-Sep-2023) Avoid ax-12 . (Revised by Rohan Ridenour, 9-Oct-2024) Use ax-pr instead of ax-pow . (Revised by BTernaryTau, 3-Dec-2024) Extract this result from the proof of dtru . (Revised by BJ, 2-Jan-2025)

Ref Expression
Assertion exneq
|- E. x -. x = y

Proof

Step Hyp Ref Expression
1 exexneq
 |-  E. z E. w -. z = w
2 equeuclr
 |-  ( w = y -> ( z = y -> z = w ) )
3 2 con3d
 |-  ( w = y -> ( -. z = w -> -. z = y ) )
4 ax7v1
 |-  ( x = z -> ( x = y -> z = y ) )
5 4 con3d
 |-  ( x = z -> ( -. z = y -> -. x = y ) )
6 5 spimevw
 |-  ( -. z = y -> E. x -. x = y )
7 3 6 syl6
 |-  ( w = y -> ( -. z = w -> E. x -. x = y ) )
8 ax7v1
 |-  ( x = w -> ( x = y -> w = y ) )
9 8 con3d
 |-  ( x = w -> ( -. w = y -> -. x = y ) )
10 9 spimevw
 |-  ( -. w = y -> E. x -. x = y )
11 10 a1d
 |-  ( -. w = y -> ( -. z = w -> E. x -. x = y ) )
12 7 11 pm2.61i
 |-  ( -. z = w -> E. x -. x = y )
13 12 exlimivv
 |-  ( E. z E. w -. z = w -> E. x -. x = y )
14 1 13 ax-mp
 |-  E. x -. x = y