Metamath Proof Explorer


Theorem z12subscl

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

Ref Expression
Assertion z12subscl ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 -s 𝐵 ) ∈ ℤs[1/2] )

Proof

Step Hyp Ref Expression
1 z12no ⊢ ( 𝐴 ∈ ℤs[1/2] → 𝐴 ∈ No )
2 z12no ⊢ ( 𝐵 ∈ ℤs[1/2] → 𝐵 ∈ No )
3 subsval ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )
4 1 2 3 syl2an ⊢ ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )
5 z12negscl ⊢ ( 𝐵 ∈ ℤs[1/2] → ( -us ‘ 𝐵 ) ∈ ℤs[1/2] )
6 z12addscl ⊢ ( ( 𝐴 ∈ ℤs[1/2] ∧ ( -us ‘ 𝐵 ) ∈ ℤs[1/2] ) → ( 𝐴 +s ( -us ‘ 𝐵 ) ) ∈ ℤs[1/2] )
7 5 6 sylan2 ⊢ ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 +s ( -us ‘ 𝐵 ) ) ∈ ℤs[1/2] )
8 4 7 eqeltrd ⊢ ( ( 𝐴 ∈ ℤs[1/2] ∧ 𝐵 ∈ ℤs[1/2] ) → ( 𝐴 -s 𝐵 ) ∈ ℤs[1/2] )