Metamath Proof Explorer


Theorem negs11

Description: Surreal negation is one-to-one. (Contributed by Scott Fenton, 3-Feb-2025)

Ref Expression
Assertion negs11 ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ ( ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) → ( -us ‘ ( -us ‘ 𝐴 ) ) = ( -us ‘ ( -us ‘ 𝐵 ) ) )
2 negnegs ⊢ ( 𝐴 ∈ No → ( -us ‘ ( -us ‘ 𝐴 ) ) = 𝐴 )
3 negnegs ⊢ ( 𝐵 ∈ No → ( -us ‘ ( -us ‘ 𝐵 ) ) = 𝐵 )
4 2 3 eqeqan12d ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( -us ‘ ( -us ‘ 𝐴 ) ) = ( -us ‘ ( -us ‘ 𝐵 ) ) ↔ 𝐴 = 𝐵 ) )
5 1 4 imbitrid ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) → 𝐴 = 𝐵 ) )
6 fveq2 ⊢ ( 𝐴 = 𝐵 → ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) )
7 5 6 impbid1 ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) ↔ 𝐴 = 𝐵 ) )