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Proof step 4 from proof int2bits-succ (previous step)

0 ≤ j + 1, 0 ≤ j ⊦ uint2bits(j + 1) = sgn(false ' + int2bits(j) +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ j ⊦ uint2bits(j + 1) = sgn(false ' + int2bits(j) +1)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;