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Proof step 16 from proof uint2bits-add-high-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int2bits-true: ⊦ int2bits(j) = true ' ↔ j = ~1;% ⊦ i < j → ¬ j < i;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ 2 ^ i < 1 ↔ i < 0;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + 0 = i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ boolvar ' ≠ [];% ⊦ sgn(false ' + mklist(false, n)) = false ';% ⊦ fillsgn(n, []) = mklist(false, n);% ⊦ sgn([]) = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;