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Proof step 9 from proof bib-on-bits (previous step)

¬ 0 < i, 8 < # int2bits(i), i ≤ 127, ~128 ≤ i ⊦ i = bits2int(lastn(8, int2bits(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
8 < # int2bits(i), ~128 ≤ i, i ≤ 0 ⊦ i = bits2int(lastn(8, int2bits(i)))

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ ¬ i < j ↔ j ≤ i;