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Proof step 7 from proof shiftzero-by-one-01 (previous step)

fillsgn(n - 1, int2bits((i - 1) / 2)) ≠ [] ∧ fillsgn(n - 1, int2bits((i - 1) / 2)).first, i < ~1, i < 0 ⊦ n = 0, bits2int(fillsgn(n - 1, int2bits((i - 1) / 2))) + 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = (i - 1) / 2 + 2 ^ n→i(# fillsgn(n, int2bits(i)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i < ~1, i < 0 ⊦ n = 0, bits2int(fillsgn(n - 1, int2bits((i - 1) / 2))) + 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = (i - 1) / 2 + 2 ^ n→i(# fillsgn(n, int2bits(i)) - 1), fillsgn(n - 1, int2bits((i - 1) / 2)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;%0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;