Theorem itgeq12sdv

Description: Equality theorem for an integral. Deduction form. General version of itgeq1d and itgeq2sdv . (Contributed by GG, 1-Sep-2025)

	Ref	Expression
Hypotheses	itgeq12sdv.1	$$⊢\phi \to A=B$$
	itgeq12sdv.2	$$⊢\phi \to C=D$$
Assertion	itgeq12sdv	$$⊢\phi \to {\int }_{A}Cdx={\int }_{B}Ddx$$

Proof

Step	Hyp	Ref	Expression
1		itgeq12sdv.1	$$⊢\phi \to A=B$$
2		itgeq12sdv.2	$$⊢\phi \to C=D$$
3	2	oveq1d	$$⊢\phi \to \frac{C}{{\mathrm{i}}^k}=\frac{D}{{\mathrm{i}}^k}$$
4	3	fveq2d	$$⊢\phi \to \Re \left(\frac{C}{{\mathrm{i}}^k}\right)=\Re \left(\frac{D}{{\mathrm{i}}^k}\right)$$
5	1	eleq2d	$$⊢\phi \to \left(x\in A\leftrightarrow x\in B\right)$$
6	5	anbi1d	$$⊢\phi \to \left(\left(x\in A\wedge 0\le y\right)\leftrightarrow \left(x\in B\wedge 0\le y\right)\right)$$
7	6	ifbid	$$⊢\phi \to if\left(\left(x\in A\wedge 0\le y\right),y,0\right)=if\left(\left(x\in B\wedge 0\le y\right),y,0\right)$$
8	4 7	csbeq12dv	$$⊢\phi \to ⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)=⦋\Re \left(\frac{D}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)$$
9	8	mpteq2dv	$$⊢\phi \to \left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)=\left(x\in ℝ⟼⦋\Re \left(\frac{D}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)$$
10	9	fveq2d	$$⊢\phi \to {\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)={\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{D}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
11	10	oveq2d	$$⊢\phi \to {\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)={\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{D}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
12	11	sumeq2sdv	$$⊢\phi \to \sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)=\sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{D}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
13		df-itg	$$⊢{\int }_{A}Cdx=\sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)$$
14		df-itg	$$⊢{\int }_{B}Ddx=\sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{D}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
15	12 13 14	3eqtr4g	$$⊢\phi \to {\int }_{A}Cdx={\int }_{B}Ddx$$