Metamath Proof Explorer


Theorem subsval

Description: The value of surreal subtraction. (Contributed by Scott Fenton, 3-Feb-2025)

Ref Expression
Assertion subsval ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 +s ( -us ‘ 𝑦 ) ) = ( 𝐴 +s ( -us ‘ 𝑦 ) ) )
2 fveq2 ⊢ ( 𝑦 = 𝐵 → ( -us ‘ 𝑦 ) = ( -us ‘ 𝐵 ) )
3 2 oveq2d ⊢ ( 𝑦 = 𝐵 → ( 𝐴 +s ( -us ‘ 𝑦 ) ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )
4 df-subs ⊢ -s = ( 𝑥 ∈ No , 𝑦 ∈ No ↦ ( 𝑥 +s ( -us ‘ 𝑦 ) ) )
5 ovex ⊢ ( 𝐴 +s ( -us ‘ 𝐵 ) ) ∈ V
6 1 3 4 5 ovmpo ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )