Metamath Proof Explorer


Theorem negsval2

Description: Surreal negation in terms of subtraction. (Contributed by Scott Fenton, 15-Apr-2025)

Ref Expression
Assertion negsval2 ( 𝐴 ∈ No → ( -us ‘ 𝐴 ) = ( 0s -s 𝐴 ) )

Proof

Step Hyp Ref Expression
1 0no ⊢ 0s ∈ No
2 subsval ⊢ ( ( 0s ∈ No ∧ 𝐴 ∈ No ) → ( 0s -s 𝐴 ) = ( 0s +s ( -us ‘ 𝐴 ) ) )
3 1 2 mpan ⊢ ( 𝐴 ∈ No → ( 0s -s 𝐴 ) = ( 0s +s ( -us ‘ 𝐴 ) ) )
4 negscl ⊢ ( 𝐴 ∈ No → ( -us ‘ 𝐴 ) ∈ No )
5 addslid ⊢ ( ( -us ‘ 𝐴 ) ∈ No → ( 0s +s ( -us ‘ 𝐴 ) ) = ( -us ‘ 𝐴 ) )
6 4 5 syl ⊢ ( 𝐴 ∈ No → ( 0s +s ( -us ‘ 𝐴 ) ) = ( -us ‘ 𝐴 ) )
7 3 6 eqtr2d ⊢ ( 𝐴 ∈ No → ( -us ‘ 𝐴 ) = ( 0s -s 𝐴 ) )