Metamath Proof Explorer


Theorem cbvitgvw2

Description: Change bound variable and domain in an integral, using implicit substitution. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypotheses cbvitgvw2.1 ⊢ ( 𝑥 = 𝑦 → 𝐶 = 𝐷 )
cbvitgvw2.2 ⊢ ( 𝑥 = 𝑦 → 𝐴 = 𝐵 )
Assertion cbvitgvw2 ∫ 𝐴 𝐶 d 𝑥 = ∫ 𝐵 𝐷 d 𝑦

Proof

Step Hyp Ref Expression
1 cbvitgvw2.1 ⊢ ( 𝑥 = 𝑦 → 𝐶 = 𝐷 )
2 cbvitgvw2.2 ⊢ ( 𝑥 = 𝑦 → 𝐴 = 𝐵 )
3 1 fvoveq1d ⊢ ( 𝑥 = 𝑦 → ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) = ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) )
4 id ⊢ ( 𝑥 = 𝑦 → 𝑥 = 𝑦 )
5 4 2 eleq12d ⊢ ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵 ) )
6 5 anbi1d ⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) ) )
7 6 ifbid ⊢ ( 𝑥 = 𝑦 → if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) = if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) )
8 3 7 csbeq12dv ⊢ ( 𝑥 = 𝑦 → ⦋ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) = ⦋ ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) )
9 8 cbvmptv ⊢ ( 𝑥 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) = ( 𝑦 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) )
10 9 fveq2i ⊢ ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) = ( ∫2 ‘ ( 𝑦 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) )
11 10 oveq2i ⊢ ( ( i ↑ 𝑡 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) ) = ( ( i ↑ 𝑡 ) · ( ∫2 ‘ ( 𝑦 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) )
12 11 sumeq2si ⊢ Σ 𝑡 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑡 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) ) = Σ 𝑡 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑡 ) · ( ∫2 ‘ ( 𝑦 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) )
13 df-itg ⊢ ∫ 𝐴 𝐶 d 𝑥 = Σ 𝑡 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑡 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑥 ∈ 𝐴 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) )
14 df-itg ⊢ ∫ 𝐵 𝐷 d 𝑦 = Σ 𝑡 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑡 ) · ( ∫2 ‘ ( 𝑦 ∈ ℝ ↦ ⦋ ( ℜ ‘ ( 𝐷 / ( i ↑ 𝑡 ) ) ) / 𝑣 ⦌ if ( ( 𝑦 ∈ 𝐵 ∧ 0 ≤ 𝑣 ) , 𝑣 , 0 ) ) ) )
15 12 13 14 3eqtr4i ⊢ ∫ 𝐴 𝐶 d 𝑥 = ∫ 𝐵 𝐷 d 𝑦