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 ⊢ ∃ x ¬ x = y

Proof

Step Hyp Ref Expression
1 exexneq ⊢ ∃ z ∃ 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 → ∃ x ¬ x = y
7 3 6 syl6 ⊢ w = y → ¬ z = w → ∃ 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 → ∃ x ¬ x = y
11 10 a1d ⊢ ¬ w = y → ¬ z = w → ∃ x ¬ x = y
12 7 11 pm2.61i ⊢ ¬ z = w → ∃ x ¬ x = y
13 12 exlimivv ⊢ ∃ z ∃ w ¬ z = w → ∃ x ¬ x = y
14 1 13 ax-mp ⊢ ∃ x ¬ x = y