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Proof step 28 from proof int2bits-neg-03 (previous step)

0 ≤ i, ~1 = 2 ^ i * 2 + j, n→i(# int2bits(~ j)) = i + 1, 0 ≤ 2 ^ i, j + 1 < 2 ^ i, 0 < 2 ^ i + j, j < 0, j < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
~1 = 2 ^ i * 2 + j, n→i(# int2bits(~ j)) = i + 1, 0 ≤ i, 0 ≤ 2 ^ i, j + 1 < 2 ^ i, 0 < 2 ^ i + j, j < 0, j < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, j < i ⊦

Rule argument

Additional information

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