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Proof step 75 from proof int2bits-add-high (previous step)

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, ¬ fillsgn(i→n(i), int2bits(j0).rest).first ∧ (i ≠ 0 ∨ int2bits(j0).rest ≠ [])

User applied rule case distinction

And obtained the following new goals

Proof step 76:
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 ⊦ ¬ fillsgn(i→n(i), int2bits(j0).rest).first, 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

pos: right 4

Additional information

Used simplifier rules: ⊦ fillsgn(0, bits) = bits;% ⊦ 1 * i = i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;