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Proof step 20 from proof div-eq-same-01 (previous step)

0 = i0 * i + j0, i < 0, j0 < i0, 0 < i0, j0 ≤ 0, j0 ≤ i0, abs(j0) ≤ i0 ⊦ 0 = i0, ~1 = i0

User applied rule cut formula

And obtained the following new goals

Proof step 21:
0 = i0 * i + j0, 0 < i0, j0 < i0, i < 0, abs(j0) ≤ i0, j0 ≤ i0, j0 ≤ 0 ⊦ ~1 = i0, 0 = i0, j0 = 0

Rule argument

fma: j0 = 0

Additional information

Used simplifier rules: ⊦ 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 = 0 ↔ i = 0 ∨ j = 0; (ws)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s) Used formulas: j0 = 0