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Proof step 10 from proof shiftright-is-le (previous step)

0 ≤ i ∧ j ≠ 0, abs(i) < i / 2 >> (abs(j) - 1), ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i < i / 2 >> (abs(j) - 1), 0 ≤ i, 0 < i / 2 >> (abs(j) - 1), ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ 0 ≤ i → abs(i) = i;