Metamath Proof Explorer


Theorem lswccatn0lsw

Description: The last symbol of a word concatenated with a nonempty word is the last symbol of the nonempty word. (Contributed by AV, 22-Oct-2018) (Proof shortened by AV, 1-May-2020)

Ref Expression
Assertion lswccatn0lsw AWordVBWordVBlastSA++B=lastSB

Proof

Step Hyp Ref Expression
1 ccatlen AWordVBWordVA++B=A+B
2 1 oveq1d AWordVBWordVA++B1=A+B-1
3 2 3adant3 AWordVBWordVBA++B1=A+B-1
4 lencl AWordVA0
5 4 nn0zd AWordVA
6 lennncl BWordVBB
7 simpl ABA
8 nnz BB
9 zaddcl ABA+B
10 8 9 sylan2 ABA+B
11 zre AA
12 nnrp BB+
13 ltaddrp AB+A<A+B
14 11 12 13 syl2an ABA<A+B
15 7 10 14 3jca ABAA+BA<A+B
16 5 6 15 syl2an AWordVBWordVBAA+BA<A+B
17 16 3impb AWordVBWordVBAA+BA<A+B
18 fzolb AA..^A+BAA+BA<A+B
19 17 18 sylibr AWordVBWordVBAA..^A+B
20 fzoend AA..^A+BA+B-1A..^A+B
21 19 20 syl AWordVBWordVBA+B-1A..^A+B
22 3 21 eqeltrd AWordVBWordVBA++B1A..^A+B
23 ccatval2 AWordVBWordVA++B1A..^A+BA++BA++B1=BA++B-1-A
24 22 23 syld3an3 AWordVBWordVBA++BA++B1=BA++B-1-A
25 2 oveq1d AWordVBWordVA++B-1-A=A+B-1-A
26 4 nn0cnd AWordVA
27 lencl BWordVB0
28 27 nn0cnd BWordVB
29 addcl ABA+B
30 1cnd AB1
31 simpl ABA
32 29 30 31 sub32d ABA+B-1-A=A+B-A-1
33 pncan2 ABA+B-A=B
34 33 oveq1d ABA+B-A-1=B1
35 32 34 eqtrd ABA+B-1-A=B1
36 26 28 35 syl2an AWordVBWordVA+B-1-A=B1
37 25 36 eqtrd AWordVBWordVA++B-1-A=B1
38 37 3adant3 AWordVBWordVBA++B-1-A=B1
39 38 fveq2d AWordVBWordVBBA++B-1-A=BB1
40 24 39 eqtrd AWordVBWordVBA++BA++B1=BB1
41 ovex A++BV
42 lsw A++BVlastSA++B=A++BA++B1
43 41 42 mp1i AWordVBWordVBlastSA++B=A++BA++B1
44 lsw BWordVlastSB=BB1
45 44 3ad2ant2 AWordVBWordVBlastSB=BB1
46 40 43 45 3eqtr4d AWordVBWordVBlastSA++B=lastSB