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Proof step 1 from proof cast-length (conclusion)

⊦ 0 < i → # int2bits(cast(i, j)) ≤ i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i, i→n(i) < # int2bits(cast(i, j)) ⊦

Rule argument

Additional information

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