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Proof step 6 from proof uint2bits-2pot-neg-mone (previous step)

0 < i, 0 < i, 0 < j, ∀ j, i0. i→n(i0) < i→n(i) → 0 < j ∧ 0 < i0 → uint2bits(j * 2 ^ i0 - 1) = uint2bits(j - 1) + mklist(true, i→n(i0)) ⊦ uint2bits(j * 2 ^ i -1 * 2 - 1) = uint2bits(j - 1) + mklist(true, i→n(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < i, 0 < j, ∀ j, i0. i→n(i0) < i→n(i) → 0 < j ∧ 0 < i0 → uint2bits(j * 2 ^ i0 - 1) = uint2bits(j - 1) + mklist(true, i→n(i0)) ⊦ uint2bits(2 * 2 ^ (i - 1) * j - 1) = uint2bits(j - 1) + mklist(true, i→n(i))

Rule argument

Additional information

Used simplifier rules: ⊦ i -1 = i - 1;