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Proof step 10 from proof int2bits-mod2pot (previous step)

0 ≤ j, 0 ≤ j, 0 ≤ i, j0 < 2 ^ i, 0 ≤ j0, 0 < j, 2 ^ i ≤ j * 2 ^ i + j0, 0 < j * 2 ^ i + j0, 0 < 2 ^ i ⊦ uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits((j * 2 ^ i + j0) % 2 ^ i)).rest = uint2bits(j) + fillsgn(i→n(i) + 1, int2bits(j0)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ j, 0 ≤ i, j0 < 2 ^ i, 0 ≤ j0, 0 < j, 2 ^ i ≤ j * 2 ^ i + j0, 0 < j * 2 ^ i + j0, 0 < 2 ^ i ⊦ fillsgn(i→n(i) + 1, uint2bits((j * 2 ^ i + j0) % 2 ^ i)).rest = fillsgn(i→n(i) + 1, int2bits(j0)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;