Implementing Sets with Lists

Implementing Sets with Lists
Associativity of intersection

We want to prove

S1,S2,S3:Set (S1 & (S2 & S3) = (S1 & S2) & S3) .

We use attribute coinduction to prove this is a behavorial property of the input specification.

First, we prove the attribute coherence of the input specification.

Introduce the lemma coherence(SET) .

This goal is true because coherence(SET) is in the hypothesis.

Then we prove S1,S2,S3:Set N:Nat N in S1 & (S2 & S3) = N in (S1 & S2) & S3

Quantifier elimination introduces new constants n :Nat and s1 s2 s3 :Set .

Show ( n in s1 & (s2 & s3) = n in (s1 & s2) & s3 ) by reduction.

Implementing Sets with Lists

Commutativity of intersection

We want to prove

S1,S2:Set (S1 & S2 = S2 & S1) .

We use attribute coinduction to prove this is a behavorial property of the input specification.

First, we prove the attribute coherence of the input specification.

Introduce the lemma coherence(SET) .

This goal is true because coherence(SET) is in the hypothesis.

Then we prove S1,S2:Set (N:Nat ( N in S1 & S2 = N in S2 & S1 ))

Quantifier elimination introduces new constants n :Nat and s1 s2 :Set .

Show ( n in s1 & s2 = n in s2 & s1 ) by reduction.

Implementing Sets with Lists

Idempotency of intersection

We want to prove

S1:Set (S1 & S1 = S1) .

We use attribute coinduction to prove this is a behavorial property of the input specification.

First, we prove the attribute coherence of the input specification.

Introduce the lemma coherence(SET) .

This goal is true because coherence(SET) is in the hypothesis.

Then we prove S1:Set (N:Nat ( N in S1 & S1 = N in S1 ))

Quantifier elimination introduces new constants n :Nat and s1 :Set .

Show ( n in s1 & s1 = n in s1 ) by reduction.

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