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Proof step 56 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(i % 2 ^ n→i(8) - 2 ^ n→i(8))) + 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
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))

Proof step 56 from proof bib-pos-high (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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