Proof of Theorem swrdccat3b
| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | simpl 482 | . . . 4
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → (𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉)) | 
| 2 |  | simpr 484 | . . . 4
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) | 
| 3 |  | elfzubelfz 13576 | . . . . 5
⊢ (𝑀 ∈ (0...(𝐿 + (♯‘𝐵))) → (𝐿 + (♯‘𝐵)) ∈ (0...(𝐿 + (♯‘𝐵)))) | 
| 4 | 3 | adantl 481 | . . . 4
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → (𝐿 + (♯‘𝐵)) ∈ (0...(𝐿 + (♯‘𝐵)))) | 
| 5 |  | swrdccatin2.l | . . . . . 6
⊢ 𝐿 = (♯‘𝐴) | 
| 6 | 5 | pfxccat3 14772 | . . . . 5
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝑀 ∈ (0...(𝐿 + (♯‘𝐵))) ∧ (𝐿 + (♯‘𝐵)) ∈ (0...(𝐿 + (♯‘𝐵)))) → ((𝐴 ++ 𝐵) substr 〈𝑀, (𝐿 + (♯‘𝐵))〉) = if((𝐿 + (♯‘𝐵)) ≤ 𝐿, (𝐴 substr 〈𝑀, (𝐿 + (♯‘𝐵))〉), if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿))))))) | 
| 7 | 6 | imp 406 | . . . 4
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ (𝑀 ∈ (0...(𝐿 + (♯‘𝐵))) ∧ (𝐿 + (♯‘𝐵)) ∈ (0...(𝐿 + (♯‘𝐵))))) → ((𝐴 ++ 𝐵) substr 〈𝑀, (𝐿 + (♯‘𝐵))〉) = if((𝐿 + (♯‘𝐵)) ≤ 𝐿, (𝐴 substr 〈𝑀, (𝐿 + (♯‘𝐵))〉), if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿)))))) | 
| 8 | 1, 2, 4, 7 | syl12anc 837 | . . 3
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → ((𝐴 ++ 𝐵) substr 〈𝑀, (𝐿 + (♯‘𝐵))〉) = if((𝐿 + (♯‘𝐵)) ≤ 𝐿, (𝐴 substr 〈𝑀, (𝐿 + (♯‘𝐵))〉), if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿)))))) | 
| 9 | 5 | swrdccat3blem 14777 | . . . 4
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ (𝐿 + (♯‘𝐵)) ≤ 𝐿) → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = (𝐴 substr 〈𝑀, (𝐿 + (♯‘𝐵))〉)) | 
| 10 |  | iftrue 4531 | . . . . . 6
⊢ (𝐿 ≤ 𝑀 → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉)) | 
| 11 | 10 | 3ad2ant3 1136 | . . . . 5
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ 𝐿 ≤ 𝑀) → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉)) | 
| 12 |  | lencl 14571 | . . . . . . . . . . . 12
⊢ (𝐴 ∈ Word 𝑉 → (♯‘𝐴) ∈
ℕ0) | 
| 13 | 12 | nn0cnd 12589 | . . . . . . . . . . 11
⊢ (𝐴 ∈ Word 𝑉 → (♯‘𝐴) ∈ ℂ) | 
| 14 |  | lencl 14571 | . . . . . . . . . . . 12
⊢ (𝐵 ∈ Word 𝑉 → (♯‘𝐵) ∈
ℕ0) | 
| 15 | 14 | nn0cnd 12589 | . . . . . . . . . . 11
⊢ (𝐵 ∈ Word 𝑉 → (♯‘𝐵) ∈ ℂ) | 
| 16 | 5 | eqcomi 2746 | . . . . . . . . . . . . 13
⊢
(♯‘𝐴) =
𝐿 | 
| 17 | 16 | eleq1i 2832 | . . . . . . . . . . . 12
⊢
((♯‘𝐴)
∈ ℂ ↔ 𝐿
∈ ℂ) | 
| 18 |  | pncan2 11515 | . . . . . . . . . . . 12
⊢ ((𝐿 ∈ ℂ ∧
(♯‘𝐵) ∈
ℂ) → ((𝐿 +
(♯‘𝐵)) −
𝐿) = (♯‘𝐵)) | 
| 19 | 17, 18 | sylanb 581 | . . . . . . . . . . 11
⊢
(((♯‘𝐴)
∈ ℂ ∧ (♯‘𝐵) ∈ ℂ) → ((𝐿 + (♯‘𝐵)) − 𝐿) = (♯‘𝐵)) | 
| 20 | 13, 15, 19 | syl2an 596 | . . . . . . . . . 10
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → ((𝐿 + (♯‘𝐵)) − 𝐿) = (♯‘𝐵)) | 
| 21 | 20 | eqcomd 2743 | . . . . . . . . 9
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (♯‘𝐵) = ((𝐿 + (♯‘𝐵)) − 𝐿)) | 
| 22 | 21 | adantr 480 | . . . . . . . 8
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → (♯‘𝐵) = ((𝐿 + (♯‘𝐵)) − 𝐿)) | 
| 23 | 22 | 3ad2ant1 1134 | . . . . . . 7
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ 𝐿 ≤ 𝑀) → (♯‘𝐵) = ((𝐿 + (♯‘𝐵)) − 𝐿)) | 
| 24 | 23 | opeq2d 4880 | . . . . . 6
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ 𝐿 ≤ 𝑀) → 〈(𝑀 − 𝐿), (♯‘𝐵)〉 = 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉) | 
| 25 | 24 | oveq2d 7447 | . . . . 5
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ 𝐿 ≤ 𝑀) → (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉) = (𝐵 substr 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉)) | 
| 26 | 11, 25 | eqtrd 2777 | . . . 4
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ 𝐿 ≤ 𝑀) → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = (𝐵 substr 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉)) | 
| 27 |  | iffalse 4534 | . . . . . 6
⊢ (¬
𝐿 ≤ 𝑀 → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) | 
| 28 | 27 | 3ad2ant3 1136 | . . . . 5
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) | 
| 29 | 20 | adantr 480 | . . . . . . . . 9
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → ((𝐿 + (♯‘𝐵)) − 𝐿) = (♯‘𝐵)) | 
| 30 | 29 | 3ad2ant1 1134 | . . . . . . . 8
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → ((𝐿 + (♯‘𝐵)) − 𝐿) = (♯‘𝐵)) | 
| 31 | 30 | oveq2d 7447 | . . . . . . 7
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿)) = (𝐵 prefix (♯‘𝐵))) | 
| 32 |  | pfxid 14722 | . . . . . . . . . 10
⊢ (𝐵 ∈ Word 𝑉 → (𝐵 prefix (♯‘𝐵)) = 𝐵) | 
| 33 | 32 | adantl 481 | . . . . . . . . 9
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝐵 prefix (♯‘𝐵)) = 𝐵) | 
| 34 | 33 | adantr 480 | . . . . . . . 8
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → (𝐵 prefix (♯‘𝐵)) = 𝐵) | 
| 35 | 34 | 3ad2ant1 1134 | . . . . . . 7
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → (𝐵 prefix (♯‘𝐵)) = 𝐵) | 
| 36 | 31, 35 | eqtr2d 2778 | . . . . . 6
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → 𝐵 = (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿))) | 
| 37 | 36 | oveq2d 7447 | . . . . 5
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵) = ((𝐴 substr 〈𝑀, 𝐿〉) ++ (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿)))) | 
| 38 | 28, 37 | eqtrd 2777 | . . . 4
⊢ ((((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) ∧ ¬ (𝐿 + (♯‘𝐵)) ≤ 𝐿 ∧ ¬ 𝐿 ≤ 𝑀) → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = ((𝐴 substr 〈𝑀, 𝐿〉) ++ (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿)))) | 
| 39 | 9, 26, 38 | 2if2 4581 | . . 3
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)) = if((𝐿 + (♯‘𝐵)) ≤ 𝐿, (𝐴 substr 〈𝑀, (𝐿 + (♯‘𝐵))〉), if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), ((𝐿 + (♯‘𝐵)) − 𝐿)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ (𝐵 prefix ((𝐿 + (♯‘𝐵)) − 𝐿)))))) | 
| 40 | 8, 39 | eqtr4d 2780 | . 2
⊢ (((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) ∧ 𝑀 ∈ (0...(𝐿 + (♯‘𝐵)))) → ((𝐴 ++ 𝐵) substr 〈𝑀, (𝐿 + (♯‘𝐵))〉) = if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵))) | 
| 41 | 40 | ex 412 | 1
⊢ ((𝐴 ∈ Word 𝑉 ∧ 𝐵 ∈ Word 𝑉) → (𝑀 ∈ (0...(𝐿 + (♯‘𝐵))) → ((𝐴 ++ 𝐵) substr 〈𝑀, (𝐿 + (♯‘𝐵))〉) = if(𝐿 ≤ 𝑀, (𝐵 substr 〈(𝑀 − 𝐿), (♯‘𝐵)〉), ((𝐴 substr 〈𝑀, 𝐿〉) ++ 𝐵)))) |