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Proof step 18 from proof uint2bits-succ (previous step)

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

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ 0 < i ∧ i % 2 = 0 → (i + 1) % 2 = 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i0 % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j;% ⊦ i % 2 ≠ 1 ∧ 0 < i → i % 2 = 0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + false ')+1 = bits + true;