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Proof step 3 from proof int2bits-leneq-neg (previous step)

# int2bits(j) = n, j < 0 ⊦ ¬ 2 ^ (n→i(n) - 1) + j < 0 ∧ ¬ 0 ≤ j + 2 ^ (n→i(n) - 2), n = 1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
j < 0 ⊦ # int2bits(j) = 1, ¬ 2 ^ (n→i(# int2bits(j)) - 1) + j < 0 ∧ ¬ 0 ≤ j + 2 ^ (n→i(# int2bits(j)) - 2)

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]