Metamath Proof Explorer


Theorem reusv2lem2

Description: Lemma for reusv2 . (Contributed by NM, 27-Oct-2010) (Proof shortened by Mario Carneiro, 19-Nov-2016) (Proof shortened by JJ, 7-Aug-2021)

Ref Expression
Assertion reusv2lem2 ∃! x y A x = B ∃! x y A x = B

Proof

Step Hyp Ref Expression
1 eunex ∃! x y A x = B x ¬ y A x = B
2 exnal x ¬ y A x = B ¬ x y A x = B
3 1 2 sylib ∃! x y A x = B ¬ x y A x = B
4 rzal A = y A x = B
5 4 alrimiv A = x y A x = B
6 3 5 nsyl3 A = ¬ ∃! x y A x = B
7 6 pm2.21d A = ∃! x y A x = B ∃! x y A x = B
8 simpr A ∃! x y A x = B ∃! x y A x = B
9 nfra1 y y A z = B
10 nfra1 y y A x = B
11 simpr y A z = B y A x = B x = B
12 rspa y A z = B y A z = B
13 12 adantr y A z = B y A x = B z = B
14 11 13 eqtr4d y A z = B y A x = B x = z
15 eqeq1 x = z x = B z = B
16 15 ralbidv x = z y A x = B y A z = B
17 16 biimprcd y A z = B x = z y A x = B
18 17 ad2antrr y A z = B y A x = B x = z y A x = B
19 14 18 mpd y A z = B y A x = B y A x = B
20 19 exp31 y A z = B y A x = B y A x = B
21 9 10 20 rexlimd y A z = B y A x = B y A x = B
22 21 adantl A y A z = B y A x = B y A x = B
23 r19.2z A y A x = B y A x = B
24 23 ex A y A x = B y A x = B
25 24 adantr A y A z = B y A x = B y A x = B
26 22 25 impbid A y A z = B y A x = B y A x = B
27 26 eubidv A y A z = B ∃! x y A x = B ∃! x y A x = B
28 27 ex A y A z = B ∃! x y A x = B ∃! x y A x = B
29 28 exlimdv A z y A z = B ∃! x y A x = B ∃! x y A x = B
30 euex ∃! x y A x = B x y A x = B
31 16 cbvexvw x y A x = B z y A z = B
32 30 31 sylib ∃! x y A x = B z y A z = B
33 29 32 impel A ∃! x y A x = B ∃! x y A x = B ∃! x y A x = B
34 8 33 mpbird A ∃! x y A x = B ∃! x y A x = B
35 34 ex A ∃! x y A x = B ∃! x y A x = B
36 7 35 pm2.61ine ∃! x y A x = B ∃! x y A x = B