Metamath Proof Explorer


Theorem bj-dfid2ALT

Description: Alternate version of dfid2 . (Contributed by BJ, 9-Nov-2024) (Proof modification is discouraged.) Use df-id instead to make the semantics of the construction df-opab clearer. (New usage is discouraged.)

Ref Expression
Assertion bj-dfid2ALT ⊢ I = 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 10 exbii ⊢ ∃ x ∃ y z = x y ∧ x = y ↔ ∃ x z = x x
12 id ⊢ x = u → x = u
13 12 12 opeq12d ⊢ x = u → x x = u u
14 13 eqeq2d ⊢ x = u → z = x x ↔ z = u u
15 14 exexw ⊢ ∃ x z = x x ↔ ∃ x ∃ x z = x x
16 11 15 bitri ⊢ ∃ x ∃ y z = x y ∧ x = y ↔ ∃ x ∃ x z = x x
17 tru ⊢ ⊤
18 17 biantru ⊢ z = x x ↔ z = x x ∧ ⊤
19 18 exbii ⊢ ∃ x z = x x ↔ ∃ x z = x x ∧ ⊤
20 19 exbii ⊢ ∃ x ∃ x z = x x ↔ ∃ x ∃ x z = x x ∧ ⊤
21 16 20 bitri ⊢ ∃ x ∃ y z = x y ∧ x = y ↔ ∃ x ∃ x z = x x ∧ ⊤
22 21 abbii ⊢ z | ∃ x ∃ y z = x y ∧ x = y = z | ∃ x ∃ x z = x x ∧ ⊤
23 df-opab ⊢ x y | x = y = z | ∃ x ∃ y z = x y ∧ x = y
24 df-opab ⊢ x x | ⊤ = z | ∃ x ∃ x z = x x ∧ ⊤
25 22 23 24 3eqtr4i ⊢ x y | x = y = x x | ⊤
26 1 25 eqtri ⊢ I = x x | ⊤