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Proof step 9 from proof bib-pos-high (previous step)

¬(0 < i ∧ lastn(8, int2bits(i)).first ∧ 8 ≤ # int2bits(i) ∧ 8 ≠ 0), 127 < i, 0 < i % 256, 127 < i % 256, 127 < i, 0 < i, 0 ≤ i ⊦ i % 256 = b2i(i2b(bits2int(lastn(8, int2bits(i))))) + 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ int2bits(0) = false ';% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;