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Proof step 10 from proof uint2bits-add-high-02 (previous step)

¬(fillsgn(i→n(i), int2bits(j0).rest) ≠ [] ∧ (fillsgn(i→n(i), int2bits(j0).rest).first ↔ false)), 0 ≤ j0 + 2 ^ i, 2 ^ (i - 1) + j0 < 0, 0 < i, j0 < 0 ⊦ sgn(int2bits(j0).rest) = sgn(false ' + fillsgn(i→n(i), int2bits(j0).rest))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ j0 + 2 ^ i, 2 ^ (i - 1) + j0 < 0, 0 < i, j0 < 0 ⊦ sgn(int2bits(j0).rest) = sgn(false ' + fillsgn(i→n(i), int2bits(j0).rest)), ¬ fillsgn(i→n(i), int2bits(j0).rest).first ∧ (i ≠ 0 ∨ int2bits(j0).rest ≠ [])

Rule argument

Additional information

Used simplifier rules: ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ fillsgn(n, bits) = [] ↔ bits = [] ∧ n = 0;