Metamath Proof Explorer


Theorem qsubcl

Description: Closure of subtraction of rationals. (Contributed by NM, 2-Aug-2004)

Ref Expression
Assertion qsubcl ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ) → ( 𝐴𝐵 ) ∈ ℚ )

Proof

Step Hyp Ref Expression
1 qcn ( 𝐴 ∈ ℚ → 𝐴 ∈ ℂ )
2 qcn ( 𝐵 ∈ ℚ → 𝐵 ∈ ℂ )
3 negsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴𝐵 ) )
4 1 2 3 syl2an ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴𝐵 ) )
5 qnegcl ( 𝐵 ∈ ℚ → - 𝐵 ∈ ℚ )
6 qaddcl ( ( 𝐴 ∈ ℚ ∧ - 𝐵 ∈ ℚ ) → ( 𝐴 + - 𝐵 ) ∈ ℚ )
7 5 6 sylan2 ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ) → ( 𝐴 + - 𝐵 ) ∈ ℚ )
8 4 7 eqeltrrd ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ) → ( 𝐴𝐵 ) ∈ ℚ )