Metamath Proof Explorer


Theorem lesubsd

Description: Swap subtrahends in a surreal inequality. (Contributed by Scott Fenton, 29-Jan-2026)

Ref Expression
Hypotheses lesubsd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
lesubsd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
lesubsd.3 ⊢ ( 𝜑 → 𝐶 ∈ No )
Assertion lesubsd ( 𝜑 → ( 𝐴 ≤s ( 𝐵 -s 𝐶 ) ↔ 𝐶 ≤s ( 𝐵 -s 𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 lesubsd.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 lesubsd.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 lesubsd.3 ⊢ ( 𝜑 → 𝐶 ∈ No )
4 npcans ⊢ ( ( 𝐵 ∈ No ∧ 𝐴 ∈ No ) → ( ( 𝐵 -s 𝐴 ) +s 𝐴 ) = 𝐵 )
5 2 1 4 syl2anc ⊢ ( 𝜑 → ( ( 𝐵 -s 𝐴 ) +s 𝐴 ) = 𝐵 )
6 2 1 subscld ⊢ ( 𝜑 → ( 𝐵 -s 𝐴 ) ∈ No )
7 1 6 addscomd ⊢ ( 𝜑 → ( 𝐴 +s ( 𝐵 -s 𝐴 ) ) = ( ( 𝐵 -s 𝐴 ) +s 𝐴 ) )
8 npcans ⊢ ( ( 𝐵 ∈ No ∧ 𝐶 ∈ No ) → ( ( 𝐵 -s 𝐶 ) +s 𝐶 ) = 𝐵 )
9 2 3 8 syl2anc ⊢ ( 𝜑 → ( ( 𝐵 -s 𝐶 ) +s 𝐶 ) = 𝐵 )
10 5 7 9 3eqtr4rd ⊢ ( 𝜑 → ( ( 𝐵 -s 𝐶 ) +s 𝐶 ) = ( 𝐴 +s ( 𝐵 -s 𝐴 ) ) )
11 10 breq2d ⊢ ( 𝜑 → ( ( 𝐴 +s 𝐶 ) ≤s ( ( 𝐵 -s 𝐶 ) +s 𝐶 ) ↔ ( 𝐴 +s 𝐶 ) ≤s ( 𝐴 +s ( 𝐵 -s 𝐴 ) ) ) )
12 2 3 subscld ⊢ ( 𝜑 → ( 𝐵 -s 𝐶 ) ∈ No )
13 1 12 3 leadds1d ⊢ ( 𝜑 → ( 𝐴 ≤s ( 𝐵 -s 𝐶 ) ↔ ( 𝐴 +s 𝐶 ) ≤s ( ( 𝐵 -s 𝐶 ) +s 𝐶 ) ) )
14 3 6 1 leadds2d ⊢ ( 𝜑 → ( 𝐶 ≤s ( 𝐵 -s 𝐴 ) ↔ ( 𝐴 +s 𝐶 ) ≤s ( 𝐴 +s ( 𝐵 -s 𝐴 ) ) ) )
15 11 13 14 3bitr4d ⊢ ( 𝜑 → ( 𝐴 ≤s ( 𝐵 -s 𝐶 ) ↔ 𝐶 ≤s ( 𝐵 -s 𝐴 ) ) )