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Proof step 59 from proof bib-pos-high (previous step)

¬(~128 ≤ i % 256 - 256 ∧ i % 256 - 256 ≤ 127), lastn(8, int2bits(i)).first, 8 ≤ # int2bits(i), 0 < i % 256, 127 < i % 256, 127 < i, 0 < i, 0 ≤ i ⊦ i % 256 = 256 + b2i(i2b(i % 256 - 256))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
lastn(8, int2bits(i)).first, 8 ≤ # int2bits(i), 0 < i % 256, 127 < i, 0 < i, 0 ≤ i, 383 < i % 256, 128 ≤ i % 256 ⊦ i % 256 = 256 + b2i(i2b(i % 256 - 256))

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i - j ≤ i0 ↔ i ≤ j + i0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;