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Proof step 1 from proof uint2bits-mod2pot-01 (conclusion)

⊦ 0 ≤ i ∧ 0 < j ∧ 0 ≤ j0 ∧ j0 < 2 ^ i → uint2bits(j * 2 ^ i + j0) = uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits(j0)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ j0, 0 < j, 0 ≤ i, j0 < 2 ^ i ⊦ uint2bits(j * 2 ^ i + j0) = uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits(j0)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)