Metamath Proof Explorer


Theorem cantnfvalf

Description: Lemma for cantnf . The function appearing in cantnfval is unconditionally a function. (Contributed by Mario Carneiro, 20-May-2015)

Ref Expression
Hypothesis cantnfvalf.f F = seq ω k A , z B C + 𝑜 D
Assertion cantnfvalf F : ω On

Proof

Step Hyp Ref Expression
1 cantnfvalf.f F = seq ω k A , z B C + 𝑜 D
2 1 fnseqom F Fn ω
3 nn0suc x ω x = y ω x = suc y
4 fveq2 x = F x = F
5 0ex V
6 1 seqom0g V F =
7 5 6 ax-mp F =
8 4 7 eqtrdi x = F x =
9 0elon On
10 8 9 eqeltrdi x = F x On
11 1 seqomsuc y ω F suc y = y k A , z B C + 𝑜 D F y
12 df-ov y k A , z B C + 𝑜 D F y = k A , z B C + 𝑜 D y F y
13 11 12 eqtrdi y ω F suc y = k A , z B C + 𝑜 D y F y
14 df-ov C + 𝑜 D = + 𝑜 C D
15 fnoa + 𝑜 Fn On × On
16 oacl x On y On x + 𝑜 y On
17 16 rgen2 x On y On x + 𝑜 y On
18 ffnov + 𝑜 : On × On On + 𝑜 Fn On × On x On y On x + 𝑜 y On
19 15 17 18 mpbir2an + 𝑜 : On × On On
20 19 9 f0cli + 𝑜 C D On
21 14 20 eqeltri C + 𝑜 D On
22 21 rgen2w k A z B C + 𝑜 D On
23 eqid k A , z B C + 𝑜 D = k A , z B C + 𝑜 D
24 23 fmpo k A z B C + 𝑜 D On k A , z B C + 𝑜 D : A × B On
25 22 24 mpbi k A , z B C + 𝑜 D : A × B On
26 25 9 f0cli k A , z B C + 𝑜 D y F y On
27 13 26 eqeltrdi y ω F suc y On
28 fveq2 x = suc y F x = F suc y
29 28 eleq1d x = suc y F x On F suc y On
30 27 29 syl5ibrcom y ω x = suc y F x On
31 30 rexlimiv y ω x = suc y F x On
32 10 31 jaoi x = y ω x = suc y F x On
33 3 32 syl x ω F x On
34 33 rgen x ω F x On
35 ffnfv F : ω On F Fn ω x ω F x On
36 2 34 35 mpbir2an F : ω On