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 → + s ⁡ A = + s ⁡ B ↔ A = B

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ + s ⁡ A = + s ⁡ B → + s ⁡ + s ⁡ A = + s ⁡ + s ⁡ B
2 negnegs ⊢ A ∈ No → + s ⁡ + s ⁡ A = A
3 negnegs ⊢ B ∈ No → + s ⁡ + s ⁡ B = B
4 2 3 eqeqan12d ⊢ A ∈ No ∧ B ∈ No → + s ⁡ + s ⁡ A = + s ⁡ + s ⁡ B ↔ A = B
5 1 4 imbitrid ⊢ A ∈ No ∧ B ∈ No → + s ⁡ A = + s ⁡ B → A = B
6 fveq2 ⊢ A = B → + s ⁡ A = + s ⁡ B
7 5 6 impbid1 ⊢ A ∈ No ∧ B ∈ No → + s ⁡ A = + s ⁡ B ↔ A = B