Metamath Proof Explorer


Theorem subsge0d

Description: Non-negative subtraction. (Contributed by Scott Fenton, 26-May-2025)

Ref Expression
Hypotheses subsge0d.1 ( 𝜑𝐴 No )
subsge0d.2 ( 𝜑𝐵 No )
Assertion subsge0d ( 𝜑 → ( 0s ≤s ( 𝐴 -s 𝐵 ) ↔ 𝐵 ≤s 𝐴 ) )

Proof

Step Hyp Ref Expression
1 subsge0d.1 ( 𝜑𝐴 No )
2 subsge0d.2 ( 𝜑𝐵 No )
3 0sno 0s No
4 3 a1i ( 𝜑 → 0s No )
5 1 2 subscld ( 𝜑 → ( 𝐴 -s 𝐵 ) ∈ No )
6 4 5 2 sleadd1d ( 𝜑 → ( 0s ≤s ( 𝐴 -s 𝐵 ) ↔ ( 0s +s 𝐵 ) ≤s ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) ) )
7 addslid ( 𝐵 No → ( 0s +s 𝐵 ) = 𝐵 )
8 2 7 syl ( 𝜑 → ( 0s +s 𝐵 ) = 𝐵 )
9 npcans ( ( 𝐴 No 𝐵 No ) → ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) = 𝐴 )
10 1 2 9 syl2anc ( 𝜑 → ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) = 𝐴 )
11 8 10 breq12d ( 𝜑 → ( ( 0s +s 𝐵 ) ≤s ( ( 𝐴 -s 𝐵 ) +s 𝐵 ) ↔ 𝐵 ≤s 𝐴 ) )
12 6 11 bitrd ( 𝜑 → ( 0s ≤s ( 𝐴 -s 𝐵 ) ↔ 𝐵 ≤s 𝐴 ) )