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Proof step 51 from proof int2bits-add-high (previous step)

¬(∃ i. 2 ^ i - 1 = ~ j) ∧ 0 ≤ ~ j, 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(~ j + 1)) + j, 0 ≤ (2 ^ n→i(# int2bits(~ j + 1)) + j) * 2 ^ i + j0, 0 ≤ j0 + 2 ^ i, 0 < 2 ^ n→i(# int2bits(1 - j)) + j ⊦ sgn(sgn(uint2bits(2 ^ n→i(# int2bits(~ j + 1)) + 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
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(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; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + 0 = i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;