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Proof step 6 from proof shiftright-is-le (previous step)

¬(0 ≤ i ∧ j ≠ 0), abs(i) < i >> j, ∀ 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 7:
i < 0, 0 < (i >> j) + i, ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: rightshift-zero: ⊦ i >> 0 = i;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ ¬ abs(i) < i;% ⊦ ¬ i ≤ j ↔ j < i;