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Proof step 1 from proof int2bits-rec-pos-01 (conclusion)

⊦ 1 < i → int2bits(i) = int2bits(i / 2) + (i % 2 = 1)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
1 < i, 0 < i ⊦ int2bits(i) = int2bits(i / 2) + (i % 2 = 1)'

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)