Metamath Proof Explorer


Theorem fct2relem

Description: Lemma for ftc2re . (Contributed by Thierry Arnoux, 20-Dec-2021)

Ref Expression
Hypotheses ftc2re.e 𝐸 = ( 𝐶 (,) 𝐷 )
ftc2re.a ( 𝜑𝐴𝐸 )
ftc2re.b ( 𝜑𝐵𝐸 )
Assertion fct2relem ( 𝜑 → ( 𝐴 [,] 𝐵 ) ⊆ 𝐸 )

Proof

Step Hyp Ref Expression
1 ftc2re.e 𝐸 = ( 𝐶 (,) 𝐷 )
2 ftc2re.a ( 𝜑𝐴𝐸 )
3 ftc2re.b ( 𝜑𝐵𝐸 )
4 2 1 eleqtrdi ( 𝜑𝐴 ∈ ( 𝐶 (,) 𝐷 ) )
5 eliooxr ( 𝐴 ∈ ( 𝐶 (,) 𝐷 ) → ( 𝐶 ∈ ℝ*𝐷 ∈ ℝ* ) )
6 4 5 syl ( 𝜑 → ( 𝐶 ∈ ℝ*𝐷 ∈ ℝ* ) )
7 6 simpld ( 𝜑𝐶 ∈ ℝ* )
8 6 simprd ( 𝜑𝐷 ∈ ℝ* )
9 eliooord ( 𝐴 ∈ ( 𝐶 (,) 𝐷 ) → ( 𝐶 < 𝐴𝐴 < 𝐷 ) )
10 4 9 syl ( 𝜑 → ( 𝐶 < 𝐴𝐴 < 𝐷 ) )
11 10 simpld ( 𝜑𝐶 < 𝐴 )
12 3 1 eleqtrdi ( 𝜑𝐵 ∈ ( 𝐶 (,) 𝐷 ) )
13 eliooord ( 𝐵 ∈ ( 𝐶 (,) 𝐷 ) → ( 𝐶 < 𝐵𝐵 < 𝐷 ) )
14 12 13 syl ( 𝜑 → ( 𝐶 < 𝐵𝐵 < 𝐷 ) )
15 14 simprd ( 𝜑𝐵 < 𝐷 )
16 iccssioo ( ( ( 𝐶 ∈ ℝ*𝐷 ∈ ℝ* ) ∧ ( 𝐶 < 𝐴𝐵 < 𝐷 ) ) → ( 𝐴 [,] 𝐵 ) ⊆ ( 𝐶 (,) 𝐷 ) )
17 7 8 11 15 16 syl22anc ( 𝜑 → ( 𝐴 [,] 𝐵 ) ⊆ ( 𝐶 (,) 𝐷 ) )
18 17 1 sseqtrrdi ( 𝜑 → ( 𝐴 [,] 𝐵 ) ⊆ 𝐸 )