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Proof step 11 from proof modlsp-03 (previous step)

i1 = i2 + j, i1 + j1 < abs(i1), abs(j1) < abs(i1), j * i + i1 < j * i0, i1 + j1 < j, abs(j1) < j, j * i + i1 + j1 - j1 < j * i0, 0 < j, i0 < i, i1 + j1 < abs(i2), abs(j1) < abs(i2), 0 ≤ i1 + j1, j ≤ abs(i1), j ≤ abs(i2) ⊦ i1 = j, i1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
abs(j1) < abs(i2 + j), abs(j1) < abs(i2), i2 + j1 < 0, abs(j1) < j, j * i + (i2 + j) + j1 - j1 < j * i0, i0 < i, (i2 + j) + j1 < abs(i2), (i2 + j) + j1 < abs(i2 + j), j * i + i2 + j < j * i0, (i2 + j1) + 1 < j, i2 + j1 < j, 0 < j, (i2 + j1) + 1 < abs(i2 + j), (i2 + j1) + 1 < abs(i2), i2 + j1 < abs(i2 + j), i2 + j1 < abs(i2), 0 ≤ (i2 + j) + j1, j ≤ abs(i2 + j), j ≤ abs(i2) ⊦ i2 + j = 0, i2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ i +1 = i + 1; (s)%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% ⊦ i + j < j ↔ i < 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i = i + j ↔ j = 0; (s)