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Proof step 1 from proof int2bits-2pot-neg-mone (conclusion)

⊦ 0 < j ∧ 0 < i → int2bits(j * 2 ^ i - 1) = int2bits(j - 1) + mklist(true, i→n(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < j, 0 < i ⊦ int2bits(j * 2 ^ i - 1) = int2bits(j - 1) + mklist(true, i→n(i))

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)