Theorem ccatval1

Description: Value of a symbol in the left half of a concatenated word. (Contributed by Stefan O'Rear, 15-Aug-2015) (Revised by Mario Carneiro, 22-Sep-2015) (Proof shortened by AV, 30-Apr-2020) (Revised by JJ, 18-Jan-2024)

	Ref	Expression
Assertion	ccatval1	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → ( ( 𝑆 ++ 𝑇 ) ‘ 𝐼 ) = ( 𝑆 ‘ 𝐼 ) )

Proof

Step	Hyp	Ref	Expression
1		ccatfval	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ) → ( 𝑆 ++ 𝑇 ) = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑆 ) + ( ♯ ‘ 𝑇 ) ) ) ↦ if ( 𝑥 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝑥 ) , ( 𝑇 ‘ ( 𝑥 − ( ♯ ‘ 𝑆 ) ) ) ) ) )
2	1	3adant3	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 ++ 𝑇 ) = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑆 ) + ( ♯ ‘ 𝑇 ) ) ) ↦ if ( 𝑥 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝑥 ) , ( 𝑇 ‘ ( 𝑥 − ( ♯ ‘ 𝑆 ) ) ) ) ) )
3		eleq1	⊢ ( 𝑥 = 𝐼 → ( 𝑥 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ↔ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) )
4		fveq2	⊢ ( 𝑥 = 𝐼 → ( 𝑆 ‘ 𝑥 ) = ( 𝑆 ‘ 𝐼 ) )
5		fvoveq1	⊢ ( 𝑥 = 𝐼 → ( 𝑇 ‘ ( 𝑥 − ( ♯ ‘ 𝑆 ) ) ) = ( 𝑇 ‘ ( 𝐼 − ( ♯ ‘ 𝑆 ) ) ) )
6	3 4 5	ifbieq12d	⊢ ( 𝑥 = 𝐼 → if ( 𝑥 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝑥 ) , ( 𝑇 ‘ ( 𝑥 − ( ♯ ‘ 𝑆 ) ) ) ) = if ( 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝐼 ) , ( 𝑇 ‘ ( 𝐼 − ( ♯ ‘ 𝑆 ) ) ) ) )
7		iftrue	⊢ ( 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) → if ( 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝐼 ) , ( 𝑇 ‘ ( 𝐼 − ( ♯ ‘ 𝑆 ) ) ) ) = ( 𝑆 ‘ 𝐼 ) )
8	7	3ad2ant3	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → if ( 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝐼 ) , ( 𝑇 ‘ ( 𝐼 − ( ♯ ‘ 𝑆 ) ) ) ) = ( 𝑆 ‘ 𝐼 ) )
9	6 8	sylan9eqr	⊢ ( ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) ∧ 𝑥 = 𝐼 ) → if ( 𝑥 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) , ( 𝑆 ‘ 𝑥 ) , ( 𝑇 ‘ ( 𝑥 − ( ♯ ‘ 𝑆 ) ) ) ) = ( 𝑆 ‘ 𝐼 ) )
10		id	⊢ ( 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) → 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) )
11		lencl	⊢ ( 𝑇 ∈ Word 𝐵 → ( ♯ ‘ 𝑇 ) ∈ ℕ0 )
12		elfzoext	⊢ ( ( 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ∧ ( ♯ ‘ 𝑇 ) ∈ ℕ0 ) → 𝐼 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑆 ) + ( ♯ ‘ 𝑇 ) ) ) )
13	10 11 12	syl2anr	⊢ ( ( 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → 𝐼 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑆 ) + ( ♯ ‘ 𝑇 ) ) ) )
14	13	3adant1	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → 𝐼 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑆 ) + ( ♯ ‘ 𝑇 ) ) ) )
15		fvexd	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → ( 𝑆 ‘ 𝐼 ) ∈ V )
16	2 9 14 15	fvmptd	⊢ ( ( 𝑆 ∈ Word 𝐴 ∧ 𝑇 ∈ Word 𝐵 ∧ 𝐼 ∈ ( 0 ..^ ( ♯ ‘ 𝑆 ) ) ) → ( ( 𝑆 ++ 𝑇 ) ‘ 𝐼 ) = ( 𝑆 ‘ 𝐼 ) )