Metamath Proof Explorer


Theorem euind

Description: Existential uniqueness via an indirect equality. (Contributed by NM, 11-Oct-2010)

Ref Expression
Hypotheses euind.1 B V
euind.2 x = y φ ψ
Assertion euind x y φ ψ A = B x φ ∃! z x φ z = A

Proof

Step Hyp Ref Expression
1 euind.1 B V
2 euind.2 x = y φ ψ
3 2 cbvexvw x φ y ψ
4 1 isseti z z = B
5 4 biantrur ψ z z = B ψ
6 5 exbii y ψ y z z = B ψ
7 19.41v z z = B ψ z z = B ψ
8 7 exbii y z z = B ψ y z z = B ψ
9 excom y z z = B ψ z y z = B ψ
10 6 8 9 3bitr2i y ψ z y z = B ψ
11 3 10 bitri x φ z y z = B ψ
12 eqeq2 A = B z = A z = B
13 12 imim2i φ ψ A = B φ ψ z = A z = B
14 biimpr z = A z = B z = B z = A
15 14 imim2i φ ψ z = A z = B φ ψ z = B z = A
16 an31 φ ψ z = B z = B ψ φ
17 16 imbi1i φ ψ z = B z = A z = B ψ φ z = A
18 impexp φ ψ z = B z = A φ ψ z = B z = A
19 impexp z = B ψ φ z = A z = B ψ φ z = A
20 17 18 19 3bitr3i φ ψ z = B z = A z = B ψ φ z = A
21 15 20 sylib φ ψ z = A z = B z = B ψ φ z = A
22 13 21 syl φ ψ A = B z = B ψ φ z = A
23 22 2alimi x y φ ψ A = B x y z = B ψ φ z = A
24 19.23v y z = B ψ φ z = A y z = B ψ φ z = A
25 24 albii x y z = B ψ φ z = A x y z = B ψ φ z = A
26 19.21v x y z = B ψ φ z = A y z = B ψ x φ z = A
27 25 26 bitri x y z = B ψ φ z = A y z = B ψ x φ z = A
28 23 27 sylib x y φ ψ A = B y z = B ψ x φ z = A
29 28 eximdv x y φ ψ A = B z y z = B ψ z x φ z = A
30 11 29 syl5bi x y φ ψ A = B x φ z x φ z = A
31 30 imp x y φ ψ A = B x φ z x φ z = A
32 pm4.24 φ φ φ
33 32 biimpi φ φ φ
34 anim12 φ z = A φ w = A φ φ z = A w = A
35 eqtr3 z = A w = A z = w
36 33 34 35 syl56 φ z = A φ w = A φ z = w
37 36 alanimi x φ z = A x φ w = A x φ z = w
38 19.23v x φ z = w x φ z = w
39 37 38 sylib x φ z = A x φ w = A x φ z = w
40 39 com12 x φ x φ z = A x φ w = A z = w
41 40 alrimivv x φ z w x φ z = A x φ w = A z = w
42 41 adantl x y φ ψ A = B x φ z w x φ z = A x φ w = A z = w
43 eqeq1 z = w z = A w = A
44 43 imbi2d z = w φ z = A φ w = A
45 44 albidv z = w x φ z = A x φ w = A
46 45 eu4 ∃! z x φ z = A z x φ z = A z w x φ z = A x φ w = A z = w
47 31 42 46 sylanbrc x y φ ψ A = B x φ ∃! z x φ z = A