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Proof step 4 from proof int2uint-fill (previous step)

¬(¬ fillsgn(n, int2bits(j)).first ∧ fillsgn(n, int2bits(j)) ≠ []), 0 ≤ j ⊦ bits2uint(fillsgn(n, int2bits(j))) = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ j, fillsgn(n, int2bits(j)).first ⊦ bits2uint(fillsgn(n, int2bits(j))) = j

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ fillsgn(n, bits) = [] ↔ bits = [] ∧ n = 0;