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Proof step 56 from proof int2bits-add-high (previous step)

0 < 2 ^ n→i(# int2bits(1 - j)) + j - 1, 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 < 2 ^ n→i(# int2bits(1 - j)) + j, 0 ≤ 2 ^ i * (2 ^ n→i(# int2bits(1 - j)) + j) + j0 ⊦ sgn((uint2bits(2 ^ n→i(# int2bits(1 - j)) + j - 1) + fillsgn(i→n(i), int2bits(j0).rest)).rest) = sgn(uint2bits(2 ^ n→i(# int2bits(1 - j)) + j - 1).rest) + 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 57:
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 ≤ 2 ^ i * (2 ^ n→i(# int2bits(1 - j)) + j) + j0, 1 < 2 ^ n→i(# int2bits(1 - j)) + j, 0 < 2 ^ n→i(# int2bits(1 - j)) + j ⊦ sgn((uint2bits(2 ^ n→i(# int2bits(1 - j)) + j - 1) + fillsgn(i→n(i), int2bits(j0).rest)).rest) = sgn(uint2bits(2 ^ n→i(# int2bits(1 - j)) + j - 1).rest) + fillsgn(i→n(i), int2bits(j0).rest), ∃ i0. 1 = 2 ^ i0 + j

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < j - i0 ↔ i + i0 < j;