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Proof step 6 from proof int2bits-mod2-len (previous step)

¬(0 ≤ i ∧ 2 ≠ 0), i ≤ 0, i % 2 < 0 ⊦ # int2bits(i % 2) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i % 2 < 0, i < 0 ⊦ # int2bits(i % 2) = 1, i = 0

Rule argument

Additional information

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