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Proof step 14 from proof bib-on-bits (previous step)

8 < # int2bits(i), i ≤ 0, ~1 ≤ i ⊦ i = 0, i = bits2int(lastn(8, int2bits(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
8 < # int2bits(~1) ⊦ ~1 = bits2int(lastn(8, int2bits(~1)))

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ i ≠ j → (i ≤ j ↔ i < j);