Metamath Proof Explorer


Theorem dtru

Description: At least two sets exist (or in terms of first-order logic, the universe of discourse has two or more objects). Note that we may not substitute the same variable for both x and y (as indicated by the distinct variable requirement), for otherwise we would contradict stdpc6 .

This theorem is proved directly from set theory axioms (no set theory definitions) and does not use ax-ext , ax-sep , or ax-pow . See dtruALT for a shorter proof using these axioms, and see dtruALT2 for a proof that uses ax-pow instead of ax-pr .

The proof makes use of dummy variables z and w which do not appear in the final theorem. They must be distinct from each other and from x and y . In other words, if we were to substitute x for z throughout the proof, the proof would fail. (Contributed by NM, 7-Nov-2006) Avoid ax-13 . (Revised by Gino Giotto, 5-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)

Ref Expression
Assertion dtru ¬ x x = y

Proof

Step Hyp Ref Expression
1 el w x w
2 ax-nul z x ¬ x z
3 elequ1 x = w x z w z
4 3 notbid x = w ¬ x z ¬ w z
5 4 spw x ¬ x z ¬ x z
6 2 5 eximii z ¬ x z
7 exdistrv w z x w ¬ x z w x w z ¬ x z
8 1 6 7 mpbir2an w z x w ¬ x z
9 ax9v2 w = z x w x z
10 9 com12 x w w = z x z
11 10 con3dimp x w ¬ x z ¬ w = z
12 11 2eximi w z x w ¬ x z w z ¬ w = z
13 equequ2 z = y w = z w = y
14 13 notbid z = y ¬ w = z ¬ w = y
15 ax7v1 x = w x = y w = y
16 15 con3d x = w ¬ w = y ¬ x = y
17 16 spimevw ¬ w = y x ¬ x = y
18 14 17 syl6bi z = y ¬ w = z x ¬ x = y
19 ax7v1 x = z x = y z = y
20 19 con3d x = z ¬ z = y ¬ x = y
21 20 spimevw ¬ z = y x ¬ x = y
22 21 a1d ¬ z = y ¬ w = z x ¬ x = y
23 18 22 pm2.61i ¬ w = z x ¬ x = y
24 23 exlimivv w z ¬ w = z x ¬ x = y
25 8 12 24 mp2b x ¬ x = y
26 exnal x ¬ x = y ¬ x x = y
27 25 26 mpbi ¬ x x = y