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Proof step 68 from proof int2bits-add-high (previous step)

0 < 2 ^ n→i(# int2bits(1 - j)) + j - 1, 0 < 2 ^ n→i(# int2bits(1 - j)) + j, 0 ≤ 2 ^ i * (2 ^ n→i(# int2bits(1 - j)) + j) + j0, j * 2 ^ i + j0 < 0, 0 ≤ i, 0 < 2 ^ i + j0, 2 ^ (i - 1) + j0 < 0, 0 < i, j0 < 0, j < 0, 0 ≤ j0 + 2 ^ i, 1 < 2 ^ n→i(# int2bits(1 - j)) + j ⊦ sgn(uint2bits(2 ^ n→i(# int2bits(1 - j)) + j - 1).rest + fillsgn(i→n(i), int2bits(j0).rest)) = sgn(uint2bits(2 ^ n→i(# int2bits(1 - j)) + j - 1).rest) + fillsgn(i→n(i), int2bits(j0).rest), ∀ i. 2 ^ n→i(# int2bits(1 - j)) + j - 1 ≠ 2 ^ i - 1, ∃ i0. 1 = 2 ^ i0 + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 69:
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(mklist(true, i→n(i0)) + fillsgn(i→n(i), int2bits(j0).rest)) = true ' + fillsgn(i→n(i), int2bits(j0).rest), ∃ i0. 1 = 2 ^ i0 + j

Rule argument

Additional information

Used simplifier rules: uint2bits-2pot-02: 0 < i ⊦ uint2bits(2 ^ i - 1) = false ' + mklist(true, i→n(i));% ⊦ 0 < i → sgn(mklist(boolvar, i→n(i))) = boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ i ≤ j → ¬ j < i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)%i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;% ⊦ i - j = i0 ↔ i = i0 + j;