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Proof step 17 from proof nummult-ls-03 (previous step)

j + 1 = i2 * (i3 + 1) + j0, j = i2 * (i3 + 1) + j0 - 1, abs(j0) < i2, j < 0, j0 < i2, 0 < i2, i3 * i2 + j0 < 0, i3 * i2 + j0 ≤ j, j0 ≤ 0 ⊦ j < i2 * i0 ↔ ¬ i0 ≤ i3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
j0 < i2, 0 < i2, i3 * i2 + j0 < 0, abs(j0) < i2, i2 * (i3 + 1) + j0 ≤ 0, j0 ≤ 0 ⊦ i2 * (i3 + 1) + j0 ≤ i2 * i0 ↔ ¬ i0 ≤ i3

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i * (j + 1) + j1 < i ↔ i * j + j1 < 0; (s)%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < 1 ↔ i ≤ 0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i ≤ i; (s)% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i + 1 ≤ j ↔ i < j; (s)% ⊦ i + j ≤ i0 + j ↔ i ≤ i0; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (j - i) = j; (s)