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Proof step 10 from proof div-le-same-neg (previous step)

i0 * (i1 + 1) + j0 < 0, 0 ≤ i1, abs(j0) ≤ i1, i0 ≤ 0, j0 ≤ 0 ⊦ 0 ≤ i1 * i0 + j0 ↔ i0 * (i1 + 1) + j0 = 0 ∨ 0 = i1

User applied rule cut formula

And obtained the following new goals

Proof step 11:
i0 * (i1 + 1) + j0 < 0, j0 ≤ 0, i0 ≤ 0, abs(j0) ≤ i1, 0 ≤ i1 ⊦ 0 ≤ i1 * i0 + j0 ↔ i0 * (i1 + 1) + j0 = 0 ∨ 0 = i1, i1 = 0

Rule argument

fma: i1 = 0

Additional information

Used simplifier rules: ⊦ i * 0 = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ 1 * i = i; (s)% ⊦ abs(i) ≤ 0 ↔ i = 0; (s) Used formulas: i1 = 0