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Proof step 8 from proof nummult-le-01 (previous step)

j + 1 - i = i1 * i + j0 ∧ j0 ≤ 0 ∧ abs(j0) < i ∧ j + 1 - i < 0, j < i, j < 0, 0 < i, j + 1 < i ⊦ i * i0 ≤ j ↔ i0 ≤ i1, i = j + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j + 1 = i * (i1 + 1) + j0, j0 ≤ 0, abs(j0) < i, j < i, j < 0, 0 < i, j0 < i, j + 1 < i ⊦ i * i0 ≤ j ↔ i0 ≤ i1, i = j + 1

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%add-zero: ⊦ i + 0 = i; from specification int-basic2%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs