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Proof step 80 from proof uint2bits-restsgn (previous step)

2 ^ (i - 1) < 4 * 2 ^ i, 2 + 2 ^ (i - 1) < i0 * 2, 2 ^ (i - 1) < i0 * 2, 2 * 2 ^ (i - 1) + 1 < 4 * 2 ^ i, 2 ^ (i - 1) < i0, 2 * 2 ^ (i - 1) + 1 < i0 * 2, 2 * 2 ^ (i - 1) < 2 ^ i, 0 ≤ i, 0 < i, 1 < i0, 1 < i, 0 < i0, 0 ≤ i0, 0 < 2 ^ i, 1 < 2 ^ i, i0 < 2 ^ i, 2 ^ i < i0 * 2, 0 ≤ 2 ^ i, i0 * 2 < 4 * 2 ^ i, (¬ uint2bits(2 * 2 ^ i - i0).rest.first) ∈ uint2bits(2 * 2 ^ i - i0).rest.rest, 2 * 2 ^ (i - 1) ≤ i0, 2 ^ (i - 1) + 1 < 2 ^ i, 2 ^ (i - 1) < 2 ^ i, 2 * (2 ^ (i - 1) + 1) < 4 * 2 ^ i, 3 + 2 ^ (i - 1) < 4 * 2 ^ i, ∀ j0, i. i→n(j0) < i→n(i0 * 2 + 0) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ i = 1, sgn(uint2bits(2 * 2 ^ i - i0).rest) = uint2bits(2 * 2 ^ i - i0).rest, 4 * 2 ^ i = 1 + i0 * 2, uint2bits(2 * 2 ^ i - i0).rest = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 81:
⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j

Rule argument

seq: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j subst: [i, j] → [i - 1, 2] rotate: T spec: int-pot inst: name: succ

Additional information

Used simplifier rules: ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + (j - i) = j;% ⊦ i + j < (* i1) * i ↔ j < (i1 - 1) * i;% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 < i ∧ 0 ≤ j → (i ^ j < i ^ j0 ↔ j < j0);% ⊦ ¬ i < i;% ⊦ i * j = j * i; Used terms: i - 1, 2