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Proof step 6 from proof bits2uint-zero-01 (previous step)

¬ true ∈ fillsgn(n, int2bits(i)), bits2uint(fillsgn(n, int2bits(i))) = 0, i < 0 ⊦ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < 0 ⊦ i = 0, true ∈ fillsgn(n, int2bits(i))

Rule argument

Additional information

Used simplifier rules: bits2uint-zeros: ⊦ ¬ true ∈ bits → bits2uint(bits) = 0;