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Proof step 3 from proof noeth-zero-ls (previous step)

abs(i) < i1 +1, 0 < abs(i), 0 < j, ∀ j, i. 0 < j ∧ 0 < abs(i) ∧ abs(i) < i1 → 0 < abs(i) * j ⊦ 0 < abs(i) * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < j, abs(i) * j < j, abs(i) ≤ i1, abs(i) * j ≤ 0, ∀ j0, i0. 0 < j0 ∧ i0 ≠ 0 ∧ abs(i0) < i1 → 0 < abs(i0) * j0 ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i +1 = i + 1; (s)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)