Metamath Proof Explorer


Theorem ditgeq123i

Description: Equality inference for the directed integral. General version of ditgeq12i and ditgeq3i . (Contributed by GG, 1-Sep-2025)

Ref Expression
Hypotheses ditgeq123i.1 ⊢ 𝐴 = 𝐵
ditgeq123i.2 ⊢ 𝐶 = 𝐷
ditgeq123i.3 ⊢ 𝐸 = 𝐹
Assertion ditgeq123i ⨜ [ 𝐴 → 𝐶 ] 𝐸 d 𝑥 = ⨜ [ 𝐵 → 𝐷 ] 𝐹 d 𝑥

Proof

Step Hyp Ref Expression
1 ditgeq123i.1 ⊢ 𝐴 = 𝐵
2 ditgeq123i.2 ⊢ 𝐶 = 𝐷
3 ditgeq123i.3 ⊢ 𝐸 = 𝐹
4 1 2 breq12i ⊢ ( 𝐴 ≤ 𝐶 ↔ 𝐵 ≤ 𝐷 )
5 1 2 oveq12i ⊢ ( 𝐴 (,) 𝐶 ) = ( 𝐵 (,) 𝐷 )
6 5 3 itgeq12i ⊢ ∫ ( 𝐴 (,) 𝐶 ) 𝐸 d 𝑥 = ∫ ( 𝐵 (,) 𝐷 ) 𝐹 d 𝑥
7 2 1 oveq12i ⊢ ( 𝐶 (,) 𝐴 ) = ( 𝐷 (,) 𝐵 )
8 7 3 itgeq12i ⊢ ∫ ( 𝐶 (,) 𝐴 ) 𝐸 d 𝑥 = ∫ ( 𝐷 (,) 𝐵 ) 𝐹 d 𝑥
9 8 negeqi ⊢ - ∫ ( 𝐶 (,) 𝐴 ) 𝐸 d 𝑥 = - ∫ ( 𝐷 (,) 𝐵 ) 𝐹 d 𝑥
10 4 6 9 ifbieq12i ⊢ if ( 𝐴 ≤ 𝐶 , ∫ ( 𝐴 (,) 𝐶 ) 𝐸 d 𝑥 , - ∫ ( 𝐶 (,) 𝐴 ) 𝐸 d 𝑥 ) = if ( 𝐵 ≤ 𝐷 , ∫ ( 𝐵 (,) 𝐷 ) 𝐹 d 𝑥 , - ∫ ( 𝐷 (,) 𝐵 ) 𝐹 d 𝑥 )
11 df-ditg ⊢ ⨜ [ 𝐴 → 𝐶 ] 𝐸 d 𝑥 = if ( 𝐴 ≤ 𝐶 , ∫ ( 𝐴 (,) 𝐶 ) 𝐸 d 𝑥 , - ∫ ( 𝐶 (,) 𝐴 ) 𝐸 d 𝑥 )
12 df-ditg ⊢ ⨜ [ 𝐵 → 𝐷 ] 𝐹 d 𝑥 = if ( 𝐵 ≤ 𝐷 , ∫ ( 𝐵 (,) 𝐷 ) 𝐹 d 𝑥 , - ∫ ( 𝐷 (,) 𝐵 ) 𝐹 d 𝑥 )
13 10 11 12 3eqtr4i ⊢ ⨜ [ 𝐴 → 𝐶 ] 𝐸 d 𝑥 = ⨜ [ 𝐵 → 𝐷 ] 𝐹 d 𝑥