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Proof step 5 from proof maxScale-not-negt (previous step)

i % 10 = 0 ∧ i ≠ 0, maxScale(i / 10) + 1 < 0, ∀ i0. i→n(i0) < i→n(i) → true → ¬ maxScale(i0) < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i % 10 = 0, maxScale(i / 10) < ~1, maxScale(i / 10) < 0, ∀ i0. i→n(i0) < i→n(i) → true → ¬ maxScale(i0) < 0 ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs