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Proof step 4 from proof int2bits-len-pos-02 (previous step)

0 ≤ j * 2 ^ i, 0 < j, 0 ≤ i ⊦ # uint2bits(j * 2 ^ i) = # int2bits(j) + i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < j, 0 ≤ i, 0 ≤ 2 ^ i ⊦ # uint2bits(j * 2 ^ i) = # int2bits(j) + i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);