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Proof step 3 from proof uint2bits-pred-intro (previous step)

j = i + 1, 0 < j ⊦ uint2bits(j) = sgn(false ' + uint2bits(i) +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ i ⊦ uint2bits(i + 1) = sgn(false ' + uint2bits(i) +1)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;