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Proof step 9 from proof shiftright-is-le (previous step)

¬(i >> j < 0 ∧ i < 0), i < 0, 0 < (i >> j) + i, ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦ j = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: shiftright-not-pos-01: ⊦ i < 0 → ¬ 0 ≤ i >> j;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;