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Proof step 14 from proof one-neg (previous step)

1 = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 < j ∧ 0 < 1, i0 < 0 ⊦ j = 0, j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
1 = i0 * j + j0, i0 < 0, j0 < j, i0 + 1 < j, i0 < j, i0 < j0, 0 < j, 0 ≤ j0 ⊦ j = ~1, j = 0

Rule argument

Additional information

Used simplifier rules: mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1