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Proof step 6 from proof shiftright-not-le (previous step)

0 ≤ i ∧ j ≠ 0, i ≤ i / 2 >> (abs(j) - 1), 0 < i, 0 < i / 2 >> (abs(j) - 1) ⊦ i = 0, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i ≤ i / 2 >> (abs(j) - 1), 0 < i, 0 < i / 2 >> (abs(j) - 1) ⊦ i = 0, j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≠ j → (i ≤ j ↔ i < j);