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 → 0 s ≤ 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 ⊢ 0 s ∈ No
4 3 a1i ⊢ φ → 0 s ∈ No
5 1 2 subscld ⊢ φ → A - s B ∈ No
6 4 5 2 leadds1d ⊢ φ → 0 s ≤ s A - s B ↔ 0 s + s B ≤ s A - s B + s B
7 addslid ⊢ B ∈ No → 0 s + s B = B
8 2 7 syl ⊢ φ → 0 s + 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 → 0 s + s B ≤ s A - s B + s B ↔ B ≤ s A
12 6 11 bitrd ⊢ φ → 0 s ≤ s A - s B ↔ B ≤ s A