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Proof step 1 from proof int2bits-len-pos-02 (conclusion)

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

Rule argument

Additional information

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