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Proof step 19 from proof int2bits-len-neg (previous step)

j + 2 ^ n→i(# bits) = 0, j < 2 ^ n→i(# bits), i < n→i(# bits), 0 < 2 ^ n→i(# bits), j ≤ i, j + 1 < 2 ^ i, j < i, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j, j + 2 ^ i < 0, j + 2 ^ i < i, j < ~1, j < 2 ^ i, j ≤ 0, 0 < 2 ^ i, j < 0, j < n→i(# bits), j + 1 < n→i(# bits), (j + 2 ^ i) + 1 < 2 ^ n→i(# bits), j + 2 ^ i < 2 ^ n→i(# bits), (j + 2 ^ i) + 1 < n→i(# bits), j + 2 ^ i < n→i(# bits), j + 1 < 2 ^ n→i(# bits) ⊦ bits = [], # bits = 1 + i→n(i), true ∈ bits

User applied rule cut formula

And obtained the following new goals

Proof step 20:
j = ~(2 ^ n→i(# bits)), j + 2 ^ n→i(# bits) = 0, j < 2 ^ n→i(# bits), i < n→i(# bits), 0 < 2 ^ n→i(# bits), j ≤ i, j + 1 < 2 ^ i, j < i, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j, j + 2 ^ i < 0, j + 2 ^ i < i, j < ~1, j < 2 ^ i, j ≤ 0, 0 < 2 ^ i, j < 0, j < n→i(# bits), j + 1 < n→i(# bits), (j + 2 ^ i) + 1 < 2 ^ n→i(# bits), j + 2 ^ i < 2 ^ n→i(# bits), (j + 2 ^ i) + 1 < n→i(# bits), j + 2 ^ i < n→i(# bits), j + 1 < 2 ^ n→i(# bits) ⊦ bits = [], # bits = 1 + i→n(i), true ∈ bits

Rule argument

fma: j = ~(2 ^ n→i(# bits))

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ (* i) ≤ j ∧ i ≠ 0 → i - 1 < j; (forward) Used formulas: j = ~(2 ^ n→i(# bits))