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Proof step 3 from proof bits2uint-zero-01 (previous step)

¬ 0 ≤ i ⊦ bits2uint(fillsgn(n, int2bits(i))) = 0 ↔ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
bits2uint(fillsgn(n, int2bits(i))) = 0, i < 0 ⊦ i = 0

Rule argument

Additional information

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