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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: int2bits-len-one: ⊦ 0 ≤ i → (# int2bits(i) = 1 ↔ i = 0);%int2bits-not-empty: ⊦ int2bits(j) ≠ [];%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]