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Proof step 18 from proof uint2bits-div2pot (previous step)

¬(0 < 2 ^ (i0 + 1) ∧ 0 ≤ j ∧ j < 2 ^ (i0 + 1)), j / 2 < 2 ^ i0, 0 < j, 2 ^ (i0 + 1) ≤ j, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 → uint2bits(j0 / 2 ^ i) = butlastn(i→n(i), uint2bits(j0)) ⊦ j = 1, uint2bits(j / 2 ^ (i0 + 1)) = butlastn(i→n(i0), uint2bits(j / 2)), ~1 = i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
j / 2 < 2 ^ i0, 0 < j, 2 ^ (i0 + 1) ≤ j, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 → uint2bits(j0 / 2 ^ i) = butlastn(i→n(i), uint2bits(j0)) ⊦ j = 1, uint2bits(j / 2 ^ (i0 + 1)) = butlastn(i→n(i0), uint2bits(j / 2)), ~1 = i0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → ¬ j < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);