Metamath Proof Explorer


Theorem sublimc

Description: Subtraction of two limits. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses sublimc.f ⊢ 𝐹 = ( 𝑥 ∈ 𝐴 ↦ 𝐵 )
sublimc.2 ⊢ 𝐺 = ( 𝑥 ∈ 𝐴 ↦ 𝐶 )
sublimc.3 ⊢ 𝐻 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 − 𝐶 ) )
sublimc.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
sublimc.5 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
sublimc.6 ⊢ ( 𝜑 → 𝐸 ∈ ( 𝐹 limℂ 𝐷 ) )
sublimc.7 ⊢ ( 𝜑 → 𝐼 ∈ ( 𝐺 limℂ 𝐷 ) )
Assertion sublimc ( 𝜑 → ( 𝐸 − 𝐼 ) ∈ ( 𝐻 limℂ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 sublimc.f ⊢ 𝐹 = ( 𝑥 ∈ 𝐴 ↦ 𝐵 )
2 sublimc.2 ⊢ 𝐺 = ( 𝑥 ∈ 𝐴 ↦ 𝐶 )
3 sublimc.3 ⊢ 𝐻 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 − 𝐶 ) )
4 sublimc.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
5 sublimc.5 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
6 sublimc.6 ⊢ ( 𝜑 → 𝐸 ∈ ( 𝐹 limℂ 𝐷 ) )
7 sublimc.7 ⊢ ( 𝜑 → 𝐼 ∈ ( 𝐺 limℂ 𝐷 ) )
8 eqid ⊢ ( 𝑥 ∈ 𝐴 ↦ - 𝐶 ) = ( 𝑥 ∈ 𝐴 ↦ - 𝐶 )
9 eqid ⊢ ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 + - 𝐶 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 + - 𝐶 ) )
10 5 negcld ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → - 𝐶 ∈ ℂ )
11 2 8 5 7 neglimc ⊢ ( 𝜑 → - 𝐼 ∈ ( ( 𝑥 ∈ 𝐴 ↦ - 𝐶 ) limℂ 𝐷 ) )
12 1 8 9 4 10 6 11 addlimc ⊢ ( 𝜑 → ( 𝐸 + - 𝐼 ) ∈ ( ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 + - 𝐶 ) ) limℂ 𝐷 ) )
13 limccl ⊢ ( 𝐹 limℂ 𝐷 ) ⊆ ℂ
14 13 6 sselid ⊢ ( 𝜑 → 𝐸 ∈ ℂ )
15 limccl ⊢ ( 𝐺 limℂ 𝐷 ) ⊆ ℂ
16 15 7 sselid ⊢ ( 𝜑 → 𝐼 ∈ ℂ )
17 14 16 negsubd ⊢ ( 𝜑 → ( 𝐸 + - 𝐼 ) = ( 𝐸 − 𝐼 ) )
18 17 eqcomd ⊢ ( 𝜑 → ( 𝐸 − 𝐼 ) = ( 𝐸 + - 𝐼 ) )
19 4 5 negsubd ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝐵 + - 𝐶 ) = ( 𝐵 − 𝐶 ) )
20 19 eqcomd ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝐵 − 𝐶 ) = ( 𝐵 + - 𝐶 ) )
21 20 mpteq2dva ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 − 𝐶 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 + - 𝐶 ) ) )
22 3 21 eqtrid ⊢ ( 𝜑 → 𝐻 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 + - 𝐶 ) ) )
23 22 oveq1d ⊢ ( 𝜑 → ( 𝐻 limℂ 𝐷 ) = ( ( 𝑥 ∈ 𝐴 ↦ ( 𝐵 + - 𝐶 ) ) limℂ 𝐷 ) )
24 12 18 23 3eltr4d ⊢ ( 𝜑 → ( 𝐸 − 𝐼 ) ∈ ( 𝐻 limℂ 𝐷 ) )