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Proof step 12 from proof uint2bits-succ (previous step)

j = i * 2 + j1 ∧ 0 ≤ j1 ∧ j1 < 2 ∧ 0 < 2, j1 = 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(i) +1 + false ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
0 < i, ∀ j0. i→n(j0) < i→n(i * 2 + 1) → 0 ≤ j0 → uint2bits(j0 + 1) = sgn(false ' + uint2bits(j0) +1) ⊦ i = 0, uint2bits(i + 1) + false ' = sgn(false ' + uint2bits(i) +1 + false ')

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ (i * 2) % 2 = 0;%j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i * j + i = i * (j + 1);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;