Theorem itgeq1f

Description: Equality theorem for an integral. (Contributed by Mario Carneiro, 28-Jun-2014) Avoid axioms. (Revised by GG, 1-Sep-2025)

	Ref	Expression
Hypotheses	itgeq1f.1	$$⊢\underset{\_}{Ⅎ}xA$$
	itgeq1f.2	$$⊢\underset{\_}{Ⅎ}xB$$
Assertion	itgeq1f	$$⊢A=B\to {\int }_{A}Cdx={\int }_{B}Cdx$$

Proof

Step	Hyp	Ref	Expression
1		itgeq1f.1	$$⊢\underset{\_}{Ⅎ}xA$$
2		itgeq1f.2	$$⊢\underset{\_}{Ⅎ}xB$$
3	1 2	nfeq	$$⊢ℲxA=B$$
4		eleq2	$$⊢A=B\to \left(x\in A\leftrightarrow x\in B\right)$$
5	4	anbi1d	$$⊢A=B\to \left(\left(x\in A\wedge 0\le y\right)\leftrightarrow \left(x\in B\wedge 0\le y\right)\right)$$
6	5	ifbid	$$⊢A=B\to if\left(\left(x\in A\wedge 0\le y\right),y,0\right)=if\left(\left(x\in B\wedge 0\le y\right),y,0\right)$$
7	6	csbeq2dv	$$⊢A=B\to ⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)=⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)$$
8	7	adantr	$$⊢\left(A=B\wedge x\in ℝ\right)\to ⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)=⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)$$
9	3 8	mpteq2da	$$⊢A=B\to \left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)=\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)$$
10	9	fveq2d	$$⊢A=B\to {\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)={\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
11	10	oveq2d	$$⊢A=B\to {\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)={\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
12	11	sumeq2sdv	$$⊢A=B\to \sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in A\wedge 0\le y\right),y,0\right)\right)\right)=\sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
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	$$⊢{\int }_{B}Cdx=\sum _{k=0}^3{\mathrm{i}}^k{\int }^2\left(\left(x\in ℝ⟼⦋\Re \left(\frac{C}{{\mathrm{i}}^k}\right)/y⦌if\left(\left(x\in B\wedge 0\le y\right),y,0\right)\right)\right)$$
15	12 13 14	3eqtr4g	$$⊢A=B\to {\int }_{A}Cdx={\int }_{B}Cdx$$