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Proof step 53 from proof uint2bits-restsgn (previous step)

i0 + 2 ^ n→i(# bits0) * 2 = 2 * 2 ^ i + 1, i0 ≤ 2 * 2 ^ i, i0 < 2 ^ i * 2, 2 ^ i < 2 * i0, i0 < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, 0 ≤ i0, 0 < i0, 1 < i, 1 < i0, 0 < i, 0 ≤ i, 0 ≤ 2 ^ i, ∀ j0, i. i→n(j0) < i→n(2 * i0 + 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 ⊦ bits0 = [], 4 * 2 ^ i = 1 + 2 * i0, false ∈ bits0

User applied rule cut formula

And obtained the following new goals

Proof step 54:
2 * 2 ^ i = 2 ^ n→i(# bits0) * 2 + (i0 - 1), i0 + 2 ^ n→i(# bits0) * 2 = 2 * 2 ^ i + 1, i0 ≤ 2 * 2 ^ i, i0 < 2 ^ i * 2, 2 ^ i < 2 * i0, i0 < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, 0 ≤ i0, 0 < i0, 1 < i, 1 < i0, 0 < i, 0 ≤ i, 0 ≤ 2 ^ i, ∀ j0, i. i→n(j0) < i→n(2 * i0 + 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 ⊦ bits0 = [], 4 * 2 ^ i = 1 + 2 * i0, false ∈ bits0

Rule argument

fma: 2 * 2 ^ i = 2 ^ n→i(# bits0) * 2 + (i0 - 1)

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i - j = i0 ↔ i = i0 + j; Used formulas: 2 * 2 ^ i = 2 ^ n→i(# bits0) * 2 + (i0 - 1)