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Proof step 2 from proof uint2bits-2pot-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 < j → uint2bits(j * 2 ^ i0) = uint2bits(j) + mklist(false, i→n(i0)) ⊦ uint2bits(j * 2 ^ i) = uint2bits(j) + mklist(false, i→n(i))

Proof step 2 from proof uint2bits-2pot-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information