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Proof step 136 from proof uint2bits-restsgn (previous step)

i - n→i(# bits) = i0 - 1, 2 ^ i0 = i2 + 1, i + 1 = i0 + n→i(# bits), 0 ≤ 2 ^ (i0 - 1), 0 < 2 ^ (i0 - 1), 1 < 2 ^ (n→i(# bits) + 1) * i2, 0 < 2 ^ (n→i(# bits) + 1), ~1 ≤ i2, 0 ≤ i, 2 ^ (n→i(# bits) + 1) * i2 + 1 < 4 * 2 ^ i, 0 ≤ 2 ^ i, 0 < i, 2 ^ i ≤ 2 ^ (n→i(# bits) + 1) * i2, i2 < 2 ^ (i0 - 1), 1 < 2 ^ i, 0 < 2 ^ i, 0 < 2 ^ n→i(# bits), n→i(# bits) ≤ i + 1, i2 * 2 ^ n→i(# bits) < 2 ^ i * 2, 1 < 2 ^ (i0 - 1), 0 ≤ i2, 0 < i2, ∀ j0, i1. i→n(j0) < i→n(2 ^ (n→i(# bits) + 1) * i2 + 1) → 0 < i1 ∧ 0 ≤ j0 ∧ 2 ^ i1 < j0 ∧ j0 < 2 * 2 ^ i1 → sgn(uint2bits(2 ^ (i1 + 2) - j0).rest) = uint2bits(2 ^ (i1 + 2) - j0).rest ⊦ sgn(bits + true ') = bits + true ', 4 * 2 ^ i = 2 * (i2 * 2 ^ n→i(# bits) + 1), false ∈ bits

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i2 = 2 ^ i0 - 1

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used formulas: i2 = 2 ^ i0 - 1