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Proof step 1 from proof int2bits-len-mod2pot (conclusion)

⊦ 0 < j ∧ 0 < i → ¬ i→n(i) + 1 < # int2bits(j % 2 ^ i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i, 0 < j, i→n(i) + 1 < # int2bits(j % 2 ^ i) ⊦

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)