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Proof step 4 from proof cast-b2i-01 (previous step)

¬(0 < i ∧ ~(2 ^ (i - 1)) ≤ b2i(by) ∧ b2i(by) < 2 ^ (i - 1)), 8 ≤ i, 0 < i, 0 ≤ i ⊦ cast(i, b2i(by)) = b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
8 ≤ i, 0 < i ⊦ cast(i, b2i(by)) = b2i(by), 0 = i, ¬ 2 ^ (i - 1) ≤ b2i(by) ∧ ¬ 2 ^ (i - 1) + b2i(by) < 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;