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Proof step 19 from proof band-011x-is-unsigned (previous step)

¬ j < ~1, i→n(i) < # int2bits(j), 0 ≤ i, 0 < i, j < 0, 0 ≤ j + 2 ^ (i - 1), 0 ≤ 2 ^ i + j ⊦ fillsgn(i→n(i), int2bits(j)) = [], j + 2 ^ i = bits2int(fillsgn(i→n(i), int2bits(j))) + 2 ^ n→i(# fillsgn(i→n(i), int2bits(j)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
i→n(i) < # int2bits(~1), 0 ≤ i, 0 < i ⊦ 2 ^ i = (bits2int(fillsgn(i→n(i), int2bits(~1))) + 2 ^ n→i(# fillsgn(i→n(i), int2bits(~1)))) + 1, fillsgn(i→n(i), int2bits(~1)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j + i0 = (i + j) + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;