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Proof step 13 from proof noeth-zero-ls (previous step)

abs(i) + 1 < i1, 0 < j, abs(i) * j < j, abs(i) * j ≤ 0, ∀ j0, i0. 0 < j0 ∧ i0 ≠ 0 ∧ abs(i0) < i1 → 0 < abs(i0) * j0 ⊦ i = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: j, i

Additional information

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