Metamath Proof Explorer


Theorem cantnflem2

Description: Lemma for cantnf . (Contributed by Mario Carneiro, 28-May-2015)

Ref Expression
Hypotheses cantnfs.s S = dom A CNF B
cantnfs.a φ A On
cantnfs.b φ B On
oemapval.t T = x y | z B x z y z w B z w x w = y w
cantnf.c φ C A 𝑜 B
cantnf.s φ C ran A CNF B
cantnf.e φ C
Assertion cantnflem2 φ A On 2 𝑜 C On 1 𝑜

Proof

Step Hyp Ref Expression
1 cantnfs.s S = dom A CNF B
2 cantnfs.a φ A On
3 cantnfs.b φ B On
4 oemapval.t T = x y | z B x z y z w B z w x w = y w
5 cantnf.c φ C A 𝑜 B
6 cantnf.s φ C ran A CNF B
7 cantnf.e φ C
8 oecl A On B On A 𝑜 B On
9 2 3 8 syl2anc φ A 𝑜 B On
10 onelon A 𝑜 B On C A 𝑜 B C On
11 9 5 10 syl2anc φ C On
12 ondif1 C On 1 𝑜 C On C
13 11 7 12 sylanbrc φ C On 1 𝑜
14 13 eldifbd φ ¬ C 1 𝑜
15 ssel A 𝑜 B 1 𝑜 C A 𝑜 B C 1 𝑜
16 5 15 syl5com φ A 𝑜 B 1 𝑜 C 1 𝑜
17 14 16 mtod φ ¬ A 𝑜 B 1 𝑜
18 oe0m B On 𝑜 B = 1 𝑜 B
19 3 18 syl φ 𝑜 B = 1 𝑜 B
20 difss 1 𝑜 B 1 𝑜
21 19 20 eqsstrdi φ 𝑜 B 1 𝑜
22 oveq1 A = A 𝑜 B = 𝑜 B
23 22 sseq1d A = A 𝑜 B 1 𝑜 𝑜 B 1 𝑜
24 21 23 syl5ibrcom φ A = A 𝑜 B 1 𝑜
25 oe1m B On 1 𝑜 𝑜 B = 1 𝑜
26 eqimss 1 𝑜 𝑜 B = 1 𝑜 1 𝑜 𝑜 B 1 𝑜
27 3 25 26 3syl φ 1 𝑜 𝑜 B 1 𝑜
28 oveq1 A = 1 𝑜 A 𝑜 B = 1 𝑜 𝑜 B
29 28 sseq1d A = 1 𝑜 A 𝑜 B 1 𝑜 1 𝑜 𝑜 B 1 𝑜
30 27 29 syl5ibrcom φ A = 1 𝑜 A 𝑜 B 1 𝑜
31 24 30 jaod φ A = A = 1 𝑜 A 𝑜 B 1 𝑜
32 17 31 mtod φ ¬ A = A = 1 𝑜
33 elpri A 1 𝑜 A = A = 1 𝑜
34 df2o3 2 𝑜 = 1 𝑜
35 33 34 eleq2s A 2 𝑜 A = A = 1 𝑜
36 32 35 nsyl φ ¬ A 2 𝑜
37 2 36 eldifd φ A On 2 𝑜
38 37 13 jca φ A On 2 𝑜 C On 1 𝑜