Metamath Proof Explorer


Theorem prodrblem

Description: Lemma for prodrb . (Contributed by Scott Fenton, 4-Dec-2017)

Ref Expression
Hypotheses prodmo.1 F=kifkAB1
prodmo.2 φkAB
prodrb.3 φNM
Assertion prodrblem φANseqM×FN=seqN×F

Proof

Step Hyp Ref Expression
1 prodmo.1 F=kifkAB1
2 prodmo.2 φkAB
3 prodrb.3 φNM
4 mullid n1n=n
5 4 adantl φANn1n=n
6 1cnd φAN1
7 3 adantr φANNM
8 iftrue kAifkAB1=B
9 8 adantl φkkAifkAB1=B
10 2 adantlr φkkAB
11 9 10 eqeltrd φkkAifkAB1
12 11 ex φkkAifkAB1
13 iffalse ¬kAifkAB1=1
14 ax-1cn 1
15 13 14 eqeltrdi ¬kAifkAB1
16 12 15 pm2.61d1 φkifkAB1
17 16 1 fmptd φF:
18 uzssz M
19 18 3 sselid φN
20 17 19 ffvelcdmd φFN
21 20 adantr φANFN
22 elfzelz nMN1n
23 22 adantl φANnMN1n
24 simplr φANnMN1AN
25 19 zcnd φN
26 25 adantr φANN
27 26 adantr φANnMN1N
28 1cnd φANnMN11
29 27 28 npcand φANnMN1N-1+1=N
30 29 fveq2d φANnMN1N-1+1=N
31 24 30 sseqtrrd φANnMN1AN-1+1
32 fznuz nMN1¬nN-1+1
33 32 adantl φANnMN1¬nN-1+1
34 31 33 ssneldd φANnMN1¬nA
35 23 34 eldifd φANnMN1nA
36 fveqeq2 k=nFk=1Fn=1
37 eldifi kAk
38 eldifn kA¬kA
39 38 13 syl kAifkAB1=1
40 39 14 eqeltrdi kAifkAB1
41 1 fvmpt2 kifkAB1Fk=ifkAB1
42 37 40 41 syl2anc kAFk=ifkAB1
43 42 39 eqtrd kAFk=1
44 36 43 vtoclga nAFn=1
45 35 44 syl φANnMN1Fn=1
46 5 6 7 21 45 seqid φANseqM×FN=seqN×F