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Proof step 9 from proof noeth-modlep (previous step)

j0 + j = ~1, j = ~1 - j0, j0 < j2, 0 < j2, j < 0, i0 < i, 0 ≤ j + j2, 0 ≤ j0, j * i0 ≤ j * i + j0, j * i0 + i0 ≤ i + j * i + j0, ∀ j0, j, i, i0. j0 < abs(j) ∧ abs(j) < j2 ∧ abs(j) * i ≤ abs(j) * i0 + j0 ∧ 0 ≤ j0 → i ≤ i0 ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i0 < i, 0 < j2, j0 < j2, (~1 - j0) * i0 + i0 ≤ i + (~1 - j0) * i + j0, (~1 - j0) * i0 ≤ (~1 - j0) * i + j0, 0 ≤ j0, ∀ j0, j, i, i0. j0 < abs(j) ∧ abs(j) < j2 ∧ abs(j) * i ≤ abs(j) * i0 + j0 ∧ 0 ≤ j0 → i ≤ i0 ⊦ ~1 = j0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + 1 ≤ j ↔ i < j; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i - j - i0 = i - j + i0; (s)%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ i + (j - i0) = i + j - i0; (s)% ⊦ i + 0 = i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (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)