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Proof step 10 from proof noeth-zero-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i * j < (i - 1) * j, 0 < (i - 1) * j, 0 < i1, i * j < i, i * j < j, 0 < i, 0 < j, i * j < i1, i * j ≤ 0, i ≤ i1, ∀ j0, i0. 0 < j0 ∧ i0 ≠ 0 ∧ abs(i0) < i1 → 0 < abs(i0) * j0 ⊦ i = 1, j = 0

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%ls-forward: ⊦ i * j < i * j0 → i ≠ 0;%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ 0 < i → abs(i - 1) = abs(i) - 1; (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 = i0 ↔ i = i0 + j; (s)