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Proof step 5 from proof int2bits-rest (previous step)

¬ 0 ≤ j, int2bits(j).rest = [] ⊦ j = ~1, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
int2bits(j).rest = [], j < 0 ⊦ j = ~1, j = 0

Rule argument

Additional information

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