Metamath Proof Explorer


Theorem nelsubclem

Description: Lemma for nelsubc . (Contributed by Zhi Wang, 5-Nov-2025)

Ref Expression
Hypotheses nelsubc.b B = Base C
nelsubc.s φ S B
nelsubc.0 φ S
nelsubc.j φ J = S × S ×
nelsubc.h H = Hom 𝑓 C
Assertion nelsubclem φ J Fn S × S J cat H ¬ x S I x J x x S y S z S f x J y ψ

Proof

Step Hyp Ref Expression
1 nelsubc.b B = Base C
2 nelsubc.s φ S B
3 nelsubc.0 φ S
4 nelsubc.j φ J = S × S ×
5 nelsubc.h H = Hom 𝑓 C
6 0ex V
7 fnconstg V S × S × Fn S × S
8 6 7 ax-mp S × S × Fn S × S
9 4 fneq1d φ J Fn S × S S × S × Fn S × S
10 8 9 mpbiri φ J Fn S × S
11 4 oveqd φ p J q = p S × S × q
12 6 ovconst2 p S q S p S × S × q =
13 11 12 sylan9eq φ p S q S p J q =
14 0ss p H q
15 13 14 eqsstrdi φ p S q S p J q p H q
16 15 ralrimivva φ p S q S p J q p H q
17 5 1 homffn H Fn B × B
18 17 a1i φ H Fn B × B
19 1 fvexi B V
20 19 a1i φ B V
21 10 18 20 isssc φ J cat H S B p S q S p J q p H q
22 2 16 21 mpbir2and φ J cat H
23 4 oveqd φ x J x = x S × S × x
24 6 ovconst2 x S x S x S × S × x =
25 24 anidms x S x S × S × x =
26 23 25 sylan9eq φ x S x J x =
27 nel02 x J x = ¬ I x J x
28 26 27 syl φ x S ¬ I x J x
29 28 reximdva0 φ S x S ¬ I x J x
30 3 29 mpdan φ x S ¬ I x J x
31 rexnal x S ¬ I x J x ¬ x S I x J x
32 30 31 sylib φ ¬ x S I x J x
33 4 oveqd φ x J y = x S × S × y
34 6 ovconst2 x S y S x S × S × y =
35 33 34 sylan9eq φ x S y S x J y =
36 rzal x J y = f x J y ψ
37 35 36 syl φ x S y S f x J y ψ
38 37 ralrimivw φ x S y S z S f x J y ψ
39 38 ralrimivva φ x S y S z S f x J y ψ
40 32 39 jca φ ¬ x S I x J x x S y S z S f x J y ψ
41 10 22 40 jca32 φ J Fn S × S J cat H ¬ x S I x J x x S y S z S f x J y ψ