Metamath Proof Explorer


Theorem ditgeq3d

Description: Equality theorem for the directed integral. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses ditgeq3d.1 ( 𝜑𝐴𝐵 )
ditgeq3d.2 ( ( 𝜑𝑥 ∈ ( 𝐴 (,) 𝐵 ) ) → 𝐷 = 𝐸 )
Assertion ditgeq3d ( 𝜑 → ⨜ [ 𝐴𝐵 ] 𝐷 d 𝑥 = ⨜ [ 𝐴𝐵 ] 𝐸 d 𝑥 )

Proof

Step Hyp Ref Expression
1 ditgeq3d.1 ( 𝜑𝐴𝐵 )
2 ditgeq3d.2 ( ( 𝜑𝑥 ∈ ( 𝐴 (,) 𝐵 ) ) → 𝐷 = 𝐸 )
3 df-ditg ⨜ [ 𝐴𝐵 ] 𝐷 d 𝑥 = if ( 𝐴𝐵 , ∫ ( 𝐴 (,) 𝐵 ) 𝐷 d 𝑥 , - ∫ ( 𝐵 (,) 𝐴 ) 𝐷 d 𝑥 )
4 1 iftrued ( 𝜑 → if ( 𝐴𝐵 , ∫ ( 𝐴 (,) 𝐵 ) 𝐷 d 𝑥 , - ∫ ( 𝐵 (,) 𝐴 ) 𝐷 d 𝑥 ) = ∫ ( 𝐴 (,) 𝐵 ) 𝐷 d 𝑥 )
5 3 4 syl5eq ( 𝜑 → ⨜ [ 𝐴𝐵 ] 𝐷 d 𝑥 = ∫ ( 𝐴 (,) 𝐵 ) 𝐷 d 𝑥 )
6 2 itgeq2dv ( 𝜑 → ∫ ( 𝐴 (,) 𝐵 ) 𝐷 d 𝑥 = ∫ ( 𝐴 (,) 𝐵 ) 𝐸 d 𝑥 )
7 df-ditg ⨜ [ 𝐴𝐵 ] 𝐸 d 𝑥 = if ( 𝐴𝐵 , ∫ ( 𝐴 (,) 𝐵 ) 𝐸 d 𝑥 , - ∫ ( 𝐵 (,) 𝐴 ) 𝐸 d 𝑥 )
8 1 iftrued ( 𝜑 → if ( 𝐴𝐵 , ∫ ( 𝐴 (,) 𝐵 ) 𝐸 d 𝑥 , - ∫ ( 𝐵 (,) 𝐴 ) 𝐸 d 𝑥 ) = ∫ ( 𝐴 (,) 𝐵 ) 𝐸 d 𝑥 )
9 7 8 syl5req ( 𝜑 → ∫ ( 𝐴 (,) 𝐵 ) 𝐸 d 𝑥 = ⨜ [ 𝐴𝐵 ] 𝐸 d 𝑥 )
10 5 6 9 3eqtrd ( 𝜑 → ⨜ [ 𝐴𝐵 ] 𝐷 d 𝑥 = ⨜ [ 𝐴𝐵 ] 𝐸 d 𝑥 )