Metamath Proof Explorer


Theorem efgredlemg

Description: Lemma for efgred . (Contributed by Mario Carneiro, 4-Jun-2016)

Ref Expression
Hypotheses efgval.w W=IWordI×2𝑜
efgval.r ˙=~FGI
efgval2.m M=yI,z2𝑜y1𝑜z
efgval2.t T=vWn0v,wI×2𝑜vsplicenn⟨“wMw”⟩
efgred.d D=WxWranTx
efgred.s S=mtWordW|t0Dk1..^ttkranTtk1mm1
efgredlem.1 φadomSbdomSSa<SASa=Sba0=b0
efgredlem.2 φAdomS
efgredlem.3 φBdomS
efgredlem.4 φSA=SB
efgredlem.5 φ¬A0=B0
efgredlemb.k K=A-1-1
efgredlemb.l L=B-1-1
efgredlemb.p φP0AK
efgredlemb.q φQ0BL
efgredlemb.u φUI×2𝑜
efgredlemb.v φVI×2𝑜
efgredlemb.6 φSA=PTAKU
efgredlemb.7 φSB=QTBLV
Assertion efgredlemg φAK=BL

Proof

Step Hyp Ref Expression
1 efgval.w W=IWordI×2𝑜
2 efgval.r ˙=~FGI
3 efgval2.m M=yI,z2𝑜y1𝑜z
4 efgval2.t T=vWn0v,wI×2𝑜vsplicenn⟨“wMw”⟩
5 efgred.d D=WxWranTx
6 efgred.s S=mtWordW|t0Dk1..^ttkranTtk1mm1
7 efgredlem.1 φadomSbdomSSa<SASa=Sba0=b0
8 efgredlem.2 φAdomS
9 efgredlem.3 φBdomS
10 efgredlem.4 φSA=SB
11 efgredlem.5 φ¬A0=B0
12 efgredlemb.k K=A-1-1
13 efgredlemb.l L=B-1-1
14 efgredlemb.p φP0AK
15 efgredlemb.q φQ0BL
16 efgredlemb.u φUI×2𝑜
17 efgredlemb.v φVI×2𝑜
18 efgredlemb.6 φSA=PTAKU
19 efgredlemb.7 φSB=QTBLV
20 fviss IWordI×2𝑜WordI×2𝑜
21 1 20 eqsstri WWordI×2𝑜
22 1 2 3 4 5 6 7 8 9 10 11 12 13 efgredlemf φAKWBLW
23 22 simpld φAKW
24 21 23 sselid φAKWordI×2𝑜
25 lencl AKWordI×2𝑜AK0
26 24 25 syl φAK0
27 26 nn0cnd φAK
28 22 simprd φBLW
29 21 28 sselid φBLWordI×2𝑜
30 lencl BLWordI×2𝑜BL0
31 29 30 syl φBL0
32 31 nn0cnd φBL
33 2cnd φ2
34 1 2 3 4 5 6 7 8 9 10 11 efgredlema φA1B1
35 34 simpld φA1
36 1 2 3 4 5 6 efgsdmi AdomSA1SAranTAA-1-1
37 8 35 36 syl2anc φSAranTAA-1-1
38 12 fveq2i AK=AA-1-1
39 38 fveq2i TAK=TAA-1-1
40 39 rneqi ranTAK=ranTAA-1-1
41 37 40 eleqtrrdi φSAranTAK
42 1 2 3 4 efgtlen AKWSAranTAKSA=AK+2
43 23 41 42 syl2anc φSA=AK+2
44 34 simprd φB1
45 1 2 3 4 5 6 efgsdmi BdomSB1SBranTBB-1-1
46 9 44 45 syl2anc φSBranTBB-1-1
47 10 46 eqeltrd φSAranTBB-1-1
48 13 fveq2i BL=BB-1-1
49 48 fveq2i TBL=TBB-1-1
50 49 rneqi ranTBL=ranTBB-1-1
51 47 50 eleqtrrdi φSAranTBL
52 1 2 3 4 efgtlen BLWSAranTBLSA=BL+2
53 28 51 52 syl2anc φSA=BL+2
54 43 53 eqtr3d φAK+2=BL+2
55 27 32 33 54 addcan2ad φAK=BL