Metamath Proof Explorer


Theorem dfttc4lem1

Description: Lemma for dfttc4 . (Contributed by Matthew House, 6-Apr-2026)

Ref Expression
Hypotheses dfttc4lem1.1 B = x | y A y z y z y = z = x
dfttc4lem1.2 C V
dfttc4lem1.3 D V
Assertion dfttc4lem1 A C z C z C = z = D D B

Proof

Step Hyp Ref Expression
1 dfttc4lem1.1 B = x | y A y z y z y = z = x
2 dfttc4lem1.2 C V
3 dfttc4lem1.3 D V
4 ineq2 y = C A y = A C
5 4 neeq1d y = C A y A C
6 ineq2 y = C z y = z C
7 6 eqeq1d y = C z y = z C =
8 7 imbi1d y = C z y = z = D z C = z = D
9 8 raleqbi1dv y = C z y z y = z = D z C z C = z = D
10 5 9 anbi12d y = C A y z y z y = z = D A C z C z C = z = D
11 2 10 spcev A C z C z C = z = D y A y z y z y = z = D
12 eqeq2 x = D z = x z = D
13 12 imbi2d x = D z y = z = x z y = z = D
14 13 ralbidv x = D z y z y = z = x z y z y = z = D
15 14 anbi2d x = D A y z y z y = z = x A y z y z y = z = D
16 15 exbidv x = D y A y z y z y = z = x y A y z y z y = z = D
17 3 16 1 elab2 D B y A y z y z y = z = D
18 11 17 sylibr A C z C z C = z = D D B