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Proof step 23 from proof bib-pos-low (previous step)

0 < i ∧ ¬ lastn(8, int2bits(i)).first ∧ 8 ≤ # int2bits(i) ∧ 8 ≠ 0, 127 < i, 0 < i, i % 256 < i, 127 < i, 0 ≤ i, i % 256 ≤ 127 ⊦ b2i(i2b(i % 2 ^ n→i(8))) = i % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
i % 256 ≤ 127, 0 ≤ i, 127 < i, i % 256 < i, 0 < i, 8 ≤ # int2bits(i) ⊦ b2i(i2b(i % 256)) = i % 256, lastn(8, int2bits(i)).first

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)