Metamath Proof Explorer


Theorem pfxccatpfx2

Description: A prefix of a concatenation of two words being the first word concatenated with a prefix of the second word. (Contributed by AV, 10-May-2020)

Ref Expression
Hypotheses swrdccatin2.l 𝐿 = ( ♯ ‘ 𝐴 )
pfxccatpfx2.m 𝑀 = ( ♯ ‘ 𝐵 )
Assertion pfxccatpfx2 ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐴 ++ 𝐵 ) prefix 𝑁 ) = ( 𝐴 ++ ( 𝐵 prefix ( 𝑁𝐿 ) ) ) )

Proof

Step Hyp Ref Expression
1 swrdccatin2.l 𝐿 = ( ♯ ‘ 𝐴 )
2 pfxccatpfx2.m 𝑀 = ( ♯ ‘ 𝐵 )
3 ccatcl ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → ( 𝐴 ++ 𝐵 ) ∈ Word 𝑉 )
4 3 3adant3 ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( 𝐴 ++ 𝐵 ) ∈ Word 𝑉 )
5 lencl ( 𝐴 ∈ Word 𝑉 → ( ♯ ‘ 𝐴 ) ∈ ℕ0 )
6 1 5 eqeltrid ( 𝐴 ∈ Word 𝑉𝐿 ∈ ℕ0 )
7 elfzuz ( 𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) → 𝑁 ∈ ( ℤ ‘ ( 𝐿 + 1 ) ) )
8 peano2nn0 ( 𝐿 ∈ ℕ0 → ( 𝐿 + 1 ) ∈ ℕ0 )
9 8 anim1i ( ( 𝐿 ∈ ℕ0𝑁 ∈ ( ℤ ‘ ( 𝐿 + 1 ) ) ) → ( ( 𝐿 + 1 ) ∈ ℕ0𝑁 ∈ ( ℤ ‘ ( 𝐿 + 1 ) ) ) )
10 6 7 9 syl2an ( ( 𝐴 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐿 + 1 ) ∈ ℕ0𝑁 ∈ ( ℤ ‘ ( 𝐿 + 1 ) ) ) )
11 10 3adant2 ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐿 + 1 ) ∈ ℕ0𝑁 ∈ ( ℤ ‘ ( 𝐿 + 1 ) ) ) )
12 eluznn0 ( ( ( 𝐿 + 1 ) ∈ ℕ0𝑁 ∈ ( ℤ ‘ ( 𝐿 + 1 ) ) ) → 𝑁 ∈ ℕ0 )
13 11 12 syl ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → 𝑁 ∈ ℕ0 )
14 pfxval ( ( ( 𝐴 ++ 𝐵 ) ∈ Word 𝑉𝑁 ∈ ℕ0 ) → ( ( 𝐴 ++ 𝐵 ) prefix 𝑁 ) = ( ( 𝐴 ++ 𝐵 ) substr ⟨ 0 , 𝑁 ⟩ ) )
15 4 13 14 syl2anc ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐴 ++ 𝐵 ) prefix 𝑁 ) = ( ( 𝐴 ++ 𝐵 ) substr ⟨ 0 , 𝑁 ⟩ ) )
16 3simpa ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) )
17 6 3ad2ant1 ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → 𝐿 ∈ ℕ0 )
18 0elfz ( 𝐿 ∈ ℕ0 → 0 ∈ ( 0 ... 𝐿 ) )
19 17 18 syl ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → 0 ∈ ( 0 ... 𝐿 ) )
20 5 nn0zd ( 𝐴 ∈ Word 𝑉 → ( ♯ ‘ 𝐴 ) ∈ ℤ )
21 1 20 eqeltrid ( 𝐴 ∈ Word 𝑉𝐿 ∈ ℤ )
22 21 adantr ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → 𝐿 ∈ ℤ )
23 uzid ( 𝐿 ∈ ℤ → 𝐿 ∈ ( ℤ𝐿 ) )
24 22 23 syl ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → 𝐿 ∈ ( ℤ𝐿 ) )
25 peano2uz ( 𝐿 ∈ ( ℤ𝐿 ) → ( 𝐿 + 1 ) ∈ ( ℤ𝐿 ) )
26 fzss1 ( ( 𝐿 + 1 ) ∈ ( ℤ𝐿 ) → ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ⊆ ( 𝐿 ... ( 𝐿 + 𝑀 ) ) )
27 24 25 26 3syl ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ⊆ ( 𝐿 ... ( 𝐿 + 𝑀 ) ) )
28 2 eqcomi ( ♯ ‘ 𝐵 ) = 𝑀
29 28 oveq2i ( 𝐿 + ( ♯ ‘ 𝐵 ) ) = ( 𝐿 + 𝑀 )
30 29 oveq2i ( 𝐿 ... ( 𝐿 + ( ♯ ‘ 𝐵 ) ) ) = ( 𝐿 ... ( 𝐿 + 𝑀 ) )
31 27 30 sseqtrrdi ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ⊆ ( 𝐿 ... ( 𝐿 + ( ♯ ‘ 𝐵 ) ) ) )
32 31 sseld ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → ( 𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) → 𝑁 ∈ ( 𝐿 ... ( 𝐿 + ( ♯ ‘ 𝐵 ) ) ) ) )
33 32 3impia ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → 𝑁 ∈ ( 𝐿 ... ( 𝐿 + ( ♯ ‘ 𝐵 ) ) ) )
34 19 33 jca ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( 0 ∈ ( 0 ... 𝐿 ) ∧ 𝑁 ∈ ( 𝐿 ... ( 𝐿 + ( ♯ ‘ 𝐵 ) ) ) ) )
35 1 pfxccatin12 ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉 ) → ( ( 0 ∈ ( 0 ... 𝐿 ) ∧ 𝑁 ∈ ( 𝐿 ... ( 𝐿 + ( ♯ ‘ 𝐵 ) ) ) ) → ( ( 𝐴 ++ 𝐵 ) substr ⟨ 0 , 𝑁 ⟩ ) = ( ( 𝐴 substr ⟨ 0 , 𝐿 ⟩ ) ++ ( 𝐵 prefix ( 𝑁𝐿 ) ) ) ) )
36 16 34 35 sylc ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐴 ++ 𝐵 ) substr ⟨ 0 , 𝑁 ⟩ ) = ( ( 𝐴 substr ⟨ 0 , 𝐿 ⟩ ) ++ ( 𝐵 prefix ( 𝑁𝐿 ) ) ) )
37 1 opeq2i ⟨ 0 , 𝐿 ⟩ = ⟨ 0 , ( ♯ ‘ 𝐴 ) ⟩
38 37 oveq2i ( 𝐴 substr ⟨ 0 , 𝐿 ⟩ ) = ( 𝐴 substr ⟨ 0 , ( ♯ ‘ 𝐴 ) ⟩ )
39 pfxval ( ( 𝐴 ∈ Word 𝑉 ∧ ( ♯ ‘ 𝐴 ) ∈ ℕ0 ) → ( 𝐴 prefix ( ♯ ‘ 𝐴 ) ) = ( 𝐴 substr ⟨ 0 , ( ♯ ‘ 𝐴 ) ⟩ ) )
40 5 39 mpdan ( 𝐴 ∈ Word 𝑉 → ( 𝐴 prefix ( ♯ ‘ 𝐴 ) ) = ( 𝐴 substr ⟨ 0 , ( ♯ ‘ 𝐴 ) ⟩ ) )
41 pfxid ( 𝐴 ∈ Word 𝑉 → ( 𝐴 prefix ( ♯ ‘ 𝐴 ) ) = 𝐴 )
42 40 41 eqtr3d ( 𝐴 ∈ Word 𝑉 → ( 𝐴 substr ⟨ 0 , ( ♯ ‘ 𝐴 ) ⟩ ) = 𝐴 )
43 38 42 syl5eq ( 𝐴 ∈ Word 𝑉 → ( 𝐴 substr ⟨ 0 , 𝐿 ⟩ ) = 𝐴 )
44 43 3ad2ant1 ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( 𝐴 substr ⟨ 0 , 𝐿 ⟩ ) = 𝐴 )
45 44 oveq1d ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐴 substr ⟨ 0 , 𝐿 ⟩ ) ++ ( 𝐵 prefix ( 𝑁𝐿 ) ) ) = ( 𝐴 ++ ( 𝐵 prefix ( 𝑁𝐿 ) ) ) )
46 15 36 45 3eqtrd ( ( 𝐴 ∈ Word 𝑉𝐵 ∈ Word 𝑉𝑁 ∈ ( ( 𝐿 + 1 ) ... ( 𝐿 + 𝑀 ) ) ) → ( ( 𝐴 ++ 𝐵 ) prefix 𝑁 ) = ( 𝐴 ++ ( 𝐵 prefix ( 𝑁𝐿 ) ) ) )