Metamath Proof Explorer


Theorem dfid2

Description: Alternate definition of the identity relation. Instance of dfid3 not requiring auxiliary axioms. (Contributed by NM, 15-Mar-2007) Reduce axiom usage. (Revised by Gino Giotto, 4-Nov-2024) (Proof shortened by BJ, 5-Nov-2024)

Use df-id instead to make the semantics of the constructor df-opab clearer. (New usage is discouraged.)

Ref Expression
Assertion dfid2 I = x x | x = x

Proof

Step Hyp Ref Expression
1 df-id I = x y | x = y
2 equcomi x = y y = x
3 2 opeq2d x = y x y = x x
4 3 eqeq2d x = y z = x y z = x x
5 4 pm5.32ri z = x y x = y z = x x x = y
6 5 exbii y z = x y x = y y z = x x x = y
7 ax6evr y x = y
8 19.42v y z = x x x = y z = x x y x = y
9 7 8 mpbiran2 y z = x x x = y z = x x
10 6 9 bitri y z = x y x = y z = x x
11 eqidd z = x x x = x
12 11 pm4.71i z = x x z = x x x = x
13 10 12 bitri y z = x y x = y z = x x x = x
14 13 exbii x y z = x y x = y x z = x x x = x
15 id x = u x = u
16 15 15 opeq12d x = u x x = u u
17 16 eqeq2d x = u z = x x z = u u
18 15 15 eqeq12d x = u x = x u = u
19 17 18 anbi12d x = u z = x x x = x z = u u u = u
20 19 exexw x z = x x x = x x x z = x x x = x
21 14 20 bitri x y z = x y x = y x x z = x x x = x
22 21 abbii z | x y z = x y x = y = z | x x z = x x x = x
23 df-opab x y | x = y = z | x y z = x y x = y
24 df-opab x x | x = x = z | x x z = x x x = x
25 22 23 24 3eqtr4i x y | x = y = x x | x = x
26 1 25 eqtri I = x x | x = x