Metamath Proof Explorer


Theorem bj-csbsnlem

Description: Lemma for bj-csbsn (in this lemma, x cannot occur in A ). (Contributed by BJ, 6-Oct-2018) (New usage is discouraged.)

Ref Expression
Assertion bj-csbsnlem A / x x = A

Proof

Step Hyp Ref Expression
1 abid y y | [˙A / x]˙ y x [˙A / x]˙ y x
2 df-sbc [˙A / x]˙ y x A x | y x
3 clelab A x | y x x x = A y x
4 velsn y x y = x
5 4 anbi2i x = A y x x = A y = x
6 5 exbii x x = A y x x x = A y = x
7 eqeq2 x = A y = x y = A
8 7 pm5.32i x = A y = x x = A y = A
9 8 exbii x x = A y = x x x = A y = A
10 19.41v x x = A y = A x x = A y = A
11 simpr x x = A y = A y = A
12 eqvisset y = A A V
13 elisset A V x x = A
14 12 13 syl y = A x x = A
15 14 ancri y = A x x = A y = A
16 11 15 impbii x x = A y = A y = A
17 9 10 16 3bitri x x = A y = x y = A
18 3 6 17 3bitri A x | y x y = A
19 1 2 18 3bitri y y | [˙A / x]˙ y x y = A
20 df-csb A / x x = y | [˙A / x]˙ y x
21 20 eleq2i y A / x x y y | [˙A / x]˙ y x
22 velsn y A y = A
23 19 21 22 3bitr4i y A / x x y A
24 23 eqriv A / x x = A