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Proof step 3 from proof int2bits-mod2-len (previous step)

¬ 0 ≤ i % 2 ⊦ i ≤ 0 → # int2bits(i % 2) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i ≤ 0, i % 2 < 0 ⊦ # int2bits(i % 2) = 1

Rule argument

Additional information

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