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Proof step 7 from proof int2bits-len-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < j, 0 ≤ i, 0 ≤ 2 ^ i ⊦ # uint2bits(j) = # int2bits(j)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;