Metamath Proof Explorer


Theorem subsge0d

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

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

Proof

Step Hyp Ref Expression
1 subsge0d.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 subsge0d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 0no ⊢ 0s ∈ No
4 3 a1i ⊢ ( 𝜑 → 0s ∈ No )
5 1 2 subscld ⊢ ( 𝜑 → ( 𝐴 -s 𝐵 ) ∈ No )
6 4 5 2 leadds1d ⊢ ( 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 ) )