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Proof step 8 from proof uint2bits-succ (previous step)

1 < j + 1, 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) / 2) + ((j + 1) % 2 = 1) = sgn(false ' + (uint2bits(j / 2) + (j % 2 = 1)')+1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
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) / 2) + ((j + 1) % 2 = 1)' = sgn(false ' + (uint2bits(j / 2) + (j % 2 = 1)')+1)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ bits + boolvar = bits + boolvar ';