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Proof step 114 from proof uint2bits-restsgn (previous step)

¬ false ∈ bits ∧ 0 ≤ 2 * 2 ^ i - i0 + 1, ¬ false ∧ 2 * 2 ^ i - i0 + 1 = 2 ^ n→i(# bits) - 1, 0 < 2 ^ i, 1 < 2 ^ i, i0 < 2 ^ i, 2 ^ i ≤ i0 * 2, 0 < i, 0 ≤ 2 ^ i, i0 * 2 + 1 < 4 * 2 ^ i, 0 ≤ i, 0 ≤ i0, 0 < i0, i0 < 2 * 2 ^ i, ∀ j0, i. i→n(j0) < i→n(i0 * 2 + 1) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ 4 * 2 ^ i = 2 * (i0 + 1), sgn(bits + true ') = bits + true ', false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 115:
2 * 2 ^ i = 2 ^ n→i(# bits) + i0, i0 < 2 * 2 ^ i, 0 < i0, 0 ≤ i0, 0 ≤ i, i0 * 2 + 1 < 4 * 2 ^ i, 0 ≤ 2 ^ i, 0 < i, 2 ^ i ≤ i0 * 2, i0 < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, 0 < 2 ^ n→i(# bits), ∀ j0, i. i→n(j0) < i→n(i0 * 2 + 1) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ sgn(bits + true ') = bits + true ', 4 * 2 ^ i = 2 * (i0 + 1), false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ i < i + j ↔ 0 < j;% ⊦ i * j = j * i;% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i + 1 ≤ j ↔ i < j;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;% ⊦ i - j = i0 ↔ i = i0 + j;