Metamath Proof Explorer


Theorem npcans

Description: Cancellation law for surreal subtraction. (Contributed by Scott Fenton, 4-Feb-2025)

Ref Expression
Assertion npcans ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 subscl ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) ∈ No )
2 simpr ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → 𝐵 ∈ No )
3 1 2 addscomd ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) = ( 𝐵 +s ( 𝐴 -s 𝐵 ) ) )
4 pncan3s ⊢ ( ( 𝐵 ∈ No ∧ 𝐴 ∈ No ) → ( 𝐵 +s ( 𝐴 -s 𝐵 ) ) = 𝐴 )
5 4 ancoms ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐵 +s ( 𝐴 -s 𝐵 ) ) = 𝐴 )
6 3 5 eqtrd ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) = 𝐴 )