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Proof step 4 from proof int2bits-leading-sgn (previous step)

¬ 0 ≤ i, ¬ 0 ≤ i ↔ int2bits(i).rest.first ⊦ # int2bits(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
int2bits(i).rest.first, i < 0 ⊦ # int2bits(i) = 1

Rule argument

Additional information

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