Metamath Proof Explorer


Theorem lazsubcl

Description: The extended integers are closed under subtraction. (Contributed by AV, 25-Oct-2026)

Ref Expression
Hypothesis laznegcl.r ⊢ 𝑅 = ( Z[]l ‘ 𝑆 )
Assertion lazsubcl ( 𝑆 ∈ ℂ → ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) → ( 𝐴 − 𝐵 ) ∈ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 laznegcl.r ⊢ 𝑅 = ( Z[]l ‘ 𝑆 )
2 1 eleq2i ⊢ ( 𝐴 ∈ 𝑅 ↔ 𝐴 ∈ ( Z[]l ‘ 𝑆 ) )
3 lazcn ⊢ ( 𝑆 ∈ ℂ → ( 𝐴 ∈ ( Z[]l ‘ 𝑆 ) → 𝐴 ∈ ℂ ) )
4 3 a1dd ⊢ ( 𝑆 ∈ ℂ → ( 𝐴 ∈ ( Z[]l ‘ 𝑆 ) → ( 𝐵 ∈ 𝑅 → 𝐴 ∈ ℂ ) ) )
5 2 4 biimtrid ⊢ ( 𝑆 ∈ ℂ → ( 𝐴 ∈ 𝑅 → ( 𝐵 ∈ 𝑅 → 𝐴 ∈ ℂ ) ) )
6 5 imp32 ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) ) → 𝐴 ∈ ℂ )
7 1 eleq2i ⊢ ( 𝐵 ∈ 𝑅 ↔ 𝐵 ∈ ( Z[]l ‘ 𝑆 ) )
8 lazcn ⊢ ( 𝑆 ∈ ℂ → ( 𝐵 ∈ ( Z[]l ‘ 𝑆 ) → 𝐵 ∈ ℂ ) )
9 7 8 biimtrid ⊢ ( 𝑆 ∈ ℂ → ( 𝐵 ∈ 𝑅 → 𝐵 ∈ ℂ ) )
10 9 adantld ⊢ ( 𝑆 ∈ ℂ → ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) → 𝐵 ∈ ℂ ) )
11 10 imp ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) ) → 𝐵 ∈ ℂ )
12 6 11 negsubd ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) ) → ( 𝐴 + - 𝐵 ) = ( 𝐴 − 𝐵 ) )
13 1 laznegcl ⊢ ( 𝑆 ∈ ℂ → ( 𝐵 ∈ 𝑅 → - 𝐵 ∈ 𝑅 ) )
14 13 anim2d ⊢ ( 𝑆 ∈ ℂ → ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) → ( 𝐴 ∈ 𝑅 ∧ - 𝐵 ∈ 𝑅 ) ) )
15 14 imp ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) ) → ( 𝐴 ∈ 𝑅 ∧ - 𝐵 ∈ 𝑅 ) )
16 1 lazaddcl ⊢ ( 𝑆 ∈ ℂ → ( ( 𝐴 ∈ 𝑅 ∧ - 𝐵 ∈ 𝑅 ) → ( 𝐴 + - 𝐵 ) ∈ 𝑅 ) )
17 16 imp ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ - 𝐵 ∈ 𝑅 ) ) → ( 𝐴 + - 𝐵 ) ∈ 𝑅 )
18 15 17 syldan ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) ) → ( 𝐴 + - 𝐵 ) ∈ 𝑅 )
19 12 18 eqeltrrd ⊢ ( ( 𝑆 ∈ ℂ ∧ ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) ) → ( 𝐴 − 𝐵 ) ∈ 𝑅 )
20 19 ex ⊢ ( 𝑆 ∈ ℂ → ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑅 ) → ( 𝐴 − 𝐵 ) ∈ 𝑅 ) )