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Proof step 4 from proof uint2bits-div2pot-01 (previous step)

j % 2 ^ i = 0, 0 ≤ 2 ^ i, 0 < 2 ^ i, 0 ≤ i, j < 2 ^ i, 1 < 2 ^ i, 0 < j ⊦ uint2bits(j) = uint2bits(j / 2 ^ i) + mklist(false, i→n(i))

Heuristic var elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ 0 < j ∧ 0 < i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j, i0, j0] → [j, 2 ^ i, i0, j0] rotate: T spec: int-div inst: name: div-elim

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);% ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 < i → (i * j < i ↔ j < 1);% ⊦ i < 1 ↔ i ≤ 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%⊦ 0 < j ∧ 0 < i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j); (elim) Used terms: j, 2 ^ i, i0, j0