Metamath Proof Explorer


Theorem seqomlem2

Description: Lemma for seqom . (Contributed by Stefan O'Rear, 1-Nov-2014) (Revised by Mario Carneiro, 23-Jun-2015)

Ref Expression
Hypothesis seqomlem.a Q = rec i ω , v V suc i i F v I I
Assertion seqomlem2 Q ω Fn ω

Proof

Step Hyp Ref Expression
1 seqomlem.a Q = rec i ω , v V suc i i F v I I
2 frfnom rec i ω , v V suc i i F v I I ω Fn ω
3 1 reseq1i Q ω = rec i ω , v V suc i i F v I I ω
4 3 fneq1i Q ω Fn ω rec i ω , v V suc i i F v I I ω Fn ω
5 2 4 mpbir Q ω Fn ω
6 fvres b ω Q ω b = Q b
7 1 seqomlem1 b ω Q b = b 2 nd Q b
8 6 7 eqtrd b ω Q ω b = b 2 nd Q b
9 fvex 2 nd Q b V
10 opelxpi b ω 2 nd Q b V b 2 nd Q b ω × V
11 9 10 mpan2 b ω b 2 nd Q b ω × V
12 8 11 eqeltrd b ω Q ω b ω × V
13 12 rgen b ω Q ω b ω × V
14 ffnfv Q ω : ω ω × V Q ω Fn ω b ω Q ω b ω × V
15 5 13 14 mpbir2an Q ω : ω ω × V
16 frn Q ω : ω ω × V ran Q ω ω × V
17 15 16 ax-mp ran Q ω ω × V
18 df-br a ran Q ω b a b ran Q ω
19 fvelrnb Q ω Fn ω a b ran Q ω c ω Q ω c = a b
20 5 19 ax-mp a b ran Q ω c ω Q ω c = a b
21 fvres c ω Q ω c = Q c
22 21 eqeq1d c ω Q ω c = a b Q c = a b
23 22 rexbiia c ω Q ω c = a b c ω Q c = a b
24 18 20 23 3bitri a ran Q ω b c ω Q c = a b
25 1 seqomlem1 c ω Q c = c 2 nd Q c
26 25 adantl a ω c ω Q c = c 2 nd Q c
27 26 eqeq1d a ω c ω Q c = a b c 2 nd Q c = a b
28 vex c V
29 fvex 2 nd Q c V
30 28 29 opth1 c 2 nd Q c = a b c = a
31 27 30 syl6bi a ω c ω Q c = a b c = a
32 fveqeq2 c = a Q c = a b Q a = a b
33 32 biimpd c = a Q c = a b Q a = a b
34 31 33 syli a ω c ω Q c = a b Q a = a b
35 fveq2 Q a = a b 2 nd Q a = 2 nd a b
36 vex a V
37 vex b V
38 36 37 op2nd 2 nd a b = b
39 35 38 eqtr2di Q a = a b b = 2 nd Q a
40 34 39 syl6 a ω c ω Q c = a b b = 2 nd Q a
41 40 rexlimdva a ω c ω Q c = a b b = 2 nd Q a
42 1 seqomlem1 a ω Q a = a 2 nd Q a
43 fveqeq2 c = a Q c = a 2 nd Q a Q a = a 2 nd Q a
44 43 rspcev a ω Q a = a 2 nd Q a c ω Q c = a 2 nd Q a
45 42 44 mpdan a ω c ω Q c = a 2 nd Q a
46 opeq2 b = 2 nd Q a a b = a 2 nd Q a
47 46 eqeq2d b = 2 nd Q a Q c = a b Q c = a 2 nd Q a
48 47 rexbidv b = 2 nd Q a c ω Q c = a b c ω Q c = a 2 nd Q a
49 45 48 syl5ibrcom a ω b = 2 nd Q a c ω Q c = a b
50 41 49 impbid a ω c ω Q c = a b b = 2 nd Q a
51 24 50 syl5bb a ω a ran Q ω b b = 2 nd Q a
52 51 alrimiv a ω b a ran Q ω b b = 2 nd Q a
53 fvex 2 nd Q a V
54 eqeq2 c = 2 nd Q a b = c b = 2 nd Q a
55 54 bibi2d c = 2 nd Q a a ran Q ω b b = c a ran Q ω b b = 2 nd Q a
56 55 albidv c = 2 nd Q a b a ran Q ω b b = c b a ran Q ω b b = 2 nd Q a
57 53 56 spcev b a ran Q ω b b = 2 nd Q a c b a ran Q ω b b = c
58 52 57 syl a ω c b a ran Q ω b b = c
59 eu6 ∃! b a ran Q ω b c b a ran Q ω b b = c
60 58 59 sylibr a ω ∃! b a ran Q ω b
61 60 rgen a ω ∃! b a ran Q ω b
62 dff3 ran Q ω : ω V ran Q ω ω × V a ω ∃! b a ran Q ω b
63 17 61 62 mpbir2an ran Q ω : ω V
64 df-ima Q ω = ran Q ω
65 64 feq1i Q ω : ω V ran Q ω : ω V
66 63 65 mpbir Q ω : ω V
67 dffn2 Q ω Fn ω Q ω : ω V
68 66 67 mpbir Q ω Fn ω