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Proof step 7 from proof bits2uint-zero-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ bits ≠ [] → (boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits)

Rule argument

seq: ⊦ bits ≠ [] → (boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits) subst: [bits, n, boolvar] → [int2bits(i), n, true] rotate: F spec: bitopsaux inst: name: fill-in

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2bits-truein: ⊦ true ∈ int2bits(j) ↔ j ≠ 0; Used terms: int2bits(i), n, true