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Proof step 5 from proof uint2bits-succ (previous step)

0 < j ∧ j ≠ 1, 0 ≤ j, ∀ j0. i→n(j0) < i→n(j) → 0 ≤ j0 → uint2bits(j0 + 1) = sgn(false ' + uint2bits(j0) +1) ⊦ uint2bits(j + 1) = sgn(false ' + (uint2bits(j / 2) + (j % 2 = 1))+1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, 0 ≤ j, ∀ j0. i→n(j0) < i→n(j) → 0 ≤ j0 → uint2bits(j0 + 1) = sgn(false ' + uint2bits(j0) +1) ⊦ j = 1, uint2bits(j + 1) = sgn(false ' + (uint2bits(j / 2) + (j % 2 = 1)')+1)

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';