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Proof step 14 from proof uint2bits-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (i * j < i ↔ j < 1);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;%0 ≤ j, 0 ≤ j0 ⊦ 0 < i → abs(i * (* j) + (* j0)) = i * j + j0;% ⊦ 0 < j → abs(j) = j;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);