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Proof step 71 from proof int2bits-add-high (previous step)

¬ false ∈ mklist(true, i→n(i0)) ∧ mklist(true, i→n(i0)) ≠ [], 2 ^ n→i(# int2bits(1 - j)) + j = 2 ^ i0, 0 ≤ 2 ^ i * 2 ^ i0 + j0, 0 ≤ j0 + 2 ^ i, j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 0 ≤ i0, 0 < i0 ⊦ sgn(true ' + fillsgn(i→n(i), int2bits(j0).rest)) = true ' + fillsgn(i→n(i), int2bits(j0).rest), ∃ i0. 1 = 2 ^ i0 + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 72:
2 ^ n→i(# int2bits(1 - j)) + j = 2 ^ i0, 0 ≤ 2 ^ i * 2 ^ i0 + j0, 0 ≤ j0 + 2 ^ i, j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 0 < i0 ⊦ sgn(true ' + fillsgn(i→n(i), int2bits(j0).rest)) = true ' + fillsgn(i→n(i), int2bits(j0).rest), i0 = 0, ∃ i0. 1 = 2 ^ i0 + j

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ false ∈ mklist(true, n);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;