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Proof step 1 from proof int2bits-mod2 (conclusion)

⊦ j < 0 → fillsgn(2, int2bits(j % 2)).rest = int2bits(j % 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j < 0 ⊦ fillsgn(2, int2bits(j % 2)).rest = int2bits(j % 2)

Rule argument

Additional information

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