Metamath Proof Explorer


Theorem zs12subscl

Description: The dyadics are closed under subtraction. (Contributed by Scott Fenton, 12-Dec-2025)

Ref Expression
Assertion zs12subscl ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 -s 𝐵 ) ∈ ℤs[1/2] )

Proof

Step Hyp Ref Expression
1 zs12no ( 𝐴 ∈ ℤs[1/2] → 𝐴 No )
2 zs12no ( 𝐵 ∈ ℤs[1/2] → 𝐵 No )
3 subsval ( ( 𝐴 No 𝐵 No ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us𝐵 ) ) )
4 1 2 3 syl2an ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us𝐵 ) ) )
5 zs12negscl ( 𝐵 ∈ ℤs[1/2] → ( -us𝐵 ) ∈ ℤs[1/2] )
6 zs12addscl ( ( 𝐴 ∈ ℤs[1/2] ∧ ( -us𝐵 ) ∈ ℤs[1/2] ) → ( 𝐴 +s ( -us𝐵 ) ) ∈ ℤs[1/2] )
7 5 6 sylan2 ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 +s ( -us𝐵 ) ) ∈ ℤs[1/2] )
8 4 7 eqeltrd ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 -s 𝐵 ) ∈ ℤs[1/2] )