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Proof step 14 from proof uint2bits-div2pot (previous step)

0 < j, 2 ^ i ≤ j, 0 < i, j / 2 < 2 ^ (i - 1), ∀ 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 ^ i) = butlastn(i→n(i) - 1, uint2bits(j / 2)), i = 0

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
i = i0 + 1, 0 < j, 2 ^ i ≤ j, 0 < i, j / 2 < 2 ^ 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 ^ i) = butlastn(i→n(i) - 1, uint2bits(j / 2)), i = 0

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i0