First-Order Rewriting

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Delia KESNER IRIF, CNRS et Universit ´e Paris [email protected] www.irif.fr/˜kesner

First-Order Rewriting

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Rewriting System

Arewriting ruleis a pair of termsl7→rsuch thatlis not a variable andVar(r)⊆Var(l). Arewriting systemis a set of rewriting rules.

Given a rewriting systemR, therewriting relation→R is generated by the following rules:

l7→r∈ R θ(l)→R θ(r)

s→R t u[s]p→R u[t]p

The notation→⁺_Ris used for the transitive closure of→R and→^∗_Rfor the reflexive-transitive closure of→R.

RemarkThis definition implies that ift→R u, then there is a rewriting rulel7→r∈ R, a substitutionσ, a termv, and a positionp∈Pos(v)such thatt=v[σ(l)]pandu=v[σ(r)]p.

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Example

Rewriting system for Peano arithmetic:

0+y → y

s(x)+y → s(x+y)

0∗y → 0

s(x)∗y → (x∗y)+y Reduction sequence:

2∗3=s(s(0))∗s(s(s(0))) → [s(0)∗s(s(s(0)))]+s(s(s(0))) → [[0∗s(s(s(0)))]+s(s(s(0)))]+s(s(s(0))) → [0+s(s(s(0)))]+s(s(s(0))) →

s(s(s(0)))+s(s(s(0))) →

s(s(s(0))+s(s(s(0)))) →

s(s(s(0)+s(s(s(0))))) →

s(s(s(0+s(s(s(0)))))) →

s(s(s(s(s(s(0))))))=6

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Confluence

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Motivations

Confluence is an undecidable property.

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Decision procedures

Sound methods Terminating methods

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Some techniques to show confluence

Confluence by strong confluence Confluence by diamond property Confluence by equivalence Confluence by commutation Confluence by interpretation Confluence by critical pairs Confluence by orthogonality Confluence by decreasing diagrams

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Confluence by strong confluence

Recall thatRisstrongly confluent (SC)iff

s →R t

↓R ↓∗R

u →⁼_R v

Example

R=











f(x,x) 7→ g(x) f(x,y) 7→ g(y) g(x) 7→ f(x,a) We only check one-step diagrams.

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Theorem

IfRisstrongly confluent, thenRisconfluent.

Proof.

Lets→ⁿtands→^mu. We reason by induction onhn,miusing the lexicographic order.

Ifn=0orm=0, then the property is trivial.

Ifn>0andm>0, we haves→¹t⁰→ⁿ⁻¹tands→¹u⁰→^m−1u.

By using the hypothesis we havet⁰→^∗vetu⁰→⁼v. We thus consideru⁰→⁼vand u⁰→^m−1u, we remark thath1/0,m−1i<lexhn,mi, and thus by the i.h. there isv⁰ such thatv→^∗v⁰etu→^∗v⁰.

Now, we considert⁰→ⁿ⁻¹tandt⁰→^∗v→^∗v⁰, we remark thathn−1,ki<lexhn,mifor anyk, and thus by the i.h.tandv⁰are joinable, and this close the whole diagram.

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Confluence by diamond property

Recall thatRhas thediamond property (DP)iff

s →R t

↓R ↓R

u →_R v

This is a particular case of strongly confluence.

Corollary

IfRhas thediamond property, thenRisconfluent.

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Confluence by equivalence

Theorem

LetRandStwo rewrite systems such that→R⊆→S⊆→^∗_RandSisstrongly confluent.

ThenRisconfluent.

Proof.

If→R⊆→S⊆→^∗_R, then→^∗_R=→^∗_S.

IfSis strongly confluent, thenSis confluent.

If→^∗_R=→^∗_S, thenRis confluent iffSis confluent.

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(Famous) Example: confluence of β

Show that(λx.t)u7→βt{x\u}is confluent.

Define a relationas follows:

xx

tt⁰ λx.tλx.t⁰

tt⁰anduu⁰ t ut⁰u⁰

tt⁰anduu⁰ (λx.t)ut⁰{x\u⁰}

LetR=βand letS=. Now,

1 Show thatβ ⊆ ⊆ β^∗.

2 Show thatis strongly confluent.

3 Conclude thatβis confluent (on all the terms).

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Confluence by commutation [Hindley-Rosen]

Two systemsRandSare said tocommuteiff

s →^∗_R t

↓∗S ↓∗S

u →^∗_R v Two systemsRandSare said tostrongly commuteiff

s →R t

↓_S ↓∗S

u →⁼_R v

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Strong Commutation

Theorem

IfRandSstrongly commute, then theycommute.

Proof.

Lets→ⁿ_Rtands→^m_Su. We reason by induction onhn,miusing the lexicographic order.

Ifn=0orm=0, then the property is trivial.

Ifn>0andm>0, thens→¹_Rt⁰→ⁿ⁻¹_R t.s→¹_Su⁰→^m−1_S u. By the hypothesisu⁰→⁼_Rvandt⁰→^∗_Sv.

Now, we consideru⁰→⁼_Rvandu⁰→^m_S⁻¹u, we remark thath0/1,m−1i<lexhn,mi, so by the i.h.u→^∗_Rv⁰andv→^∗_Sv⁰.

Similarly,t⁰→ⁿ⁻¹_R tandt⁰→^∗_Sv⁰, we remark thathn−1,ki<lexhn,mifor anyk, and then the diagram closes by the i.h.

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Theorem

IfRandSare bothconfluentandcommute, thenR∪Sisconfluent.

Proof.

Lets→^∗_R∪Stands→^∗_R∪Su. We decompose both sequences ass→^∗_R→^∗_S. . .→^∗_R→^∗_St ands→^∗_R→^∗_S. . .→^∗_R→^∗_Su.

Intuitively, the steps withRare closed by confluence ofR, and the steps withSare closed by confluence ofS. The other ones by commutation.

Formally,

We show→R∪S⊆→^∗_R→^∗_S⊆→^∗_R∪S

We remark that→^∗_R→^∗_Sis strongly confluent.

We apply confluence by equivalence.

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(Famous) Example: confluence of βη

Example

(λx.t)u 7→_β t{x\u}

λx.t x 7→_η t Ifx<fv(t)

LetR=βandS=η. Now,

Show thatβis confluent (done).

Show thatηis confluent.

Show thatβandηstrongly commute.

Conclude thatβ∪ηis confluent.

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Confluence by interpretation

Theorem

LetRandSbe two relations s.t.Ris confluent and terminating. If there is a relationT on the set ofR-normal forms s.t.

1 →^∗_T ⊆ →^∗_R∪S_and

2 a→S bimpliesR(a)→^∗_T R(b)

3 T is confluent ThenR∪Sis confluent.

Proof.

Lets→^∗_R∪Stands→^∗_R∪Su.

SinceRis confluent and terminating we can computeR-normal formss→^∗_RR(s), t→^∗_RR(t), andu→^∗_RR(u).

We can then simulate the first two→^∗_R∪S-sequences byR(s)→^∗_T R(t)andR(s)→^∗_T R(u) using point 2.

We can close this divergence byR(t)→^∗_T vandR(u)→^∗_T vusing point 3.

We obtainR(t)→^∗_R∪SvandR(u)→^∗_R∪Svusing point 1.

We thus conclude sincet→^∗_RR(t)→^∗_R∪Svandu→^∗_RR(u)→^∗_R∪Sv.

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(Famous) Example: confluence of λx

(λx.t)u 7→_B t[x/u]

(t u)[x/v] 7→_x (t[x/v]u[x/v])

(λy.t)[x/v] 7→x λy.t[x/v] ifx,y&y<fv(v)

y[x/v] 7→x y ifx,y

x[x/v] 7→x v LetR=xandS=BandT =β, Now,

Show thatxis confluent and terminating.

Show thatβ⊆(x∪B)^∗.

Show thata→_B bimpliesx(a)→^∗_βx(b). Sinceβis confluentλx=x∪Bis confluent.

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Confluence by critical pairs

Lemma

(Newmann)LetR_{be a}_SNsystem. ThenR_islocally confluentiffR_is_confluent.

Proof.

(By Huet) By well-founded induction ons∈S N.

s → t⁰ →^∗ t

↓ ↓∗ ↓∗

u⁰ →^∗ v i.h. sincet⁰<s ↓∗

↓∗ i.h. sinceu⁰<s ↓∗ ↓∗

u →^∗ p →^∗ p⁰

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Important remark

The following (infinite) system on natural numbers:

R=











2.n 7→ 2.n+1

2.n 7→ a

2.m+1 7→ 2.m+2

2.m+1 7→ b

islocally confluentbut notconfluent:a←0→^∗b In fact it is not SN

0→ 1→ 2→ 3→ . . .

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Towards local confluence: critical pairs

Acritical pairbetween twovariable disjoint rulesl7→randg7→dofR(not necessarily distinct rules) is a pairhσ(r), σ(l)[σ(d)]pis.t.

1 p∈Pos(l)andl|pis not a variable.

2 σis a principal unifier ofl|pandg.

3 Ifl7→randg7→dare the same rule, thenp,Λ.

Observe that

σ(r)←σ(l) =σ(l)[σ(l)|p]_p

=σ(l)[σ(l|p)]p

=σ(l)[σ(g)]p→ σ(l)[σ(d)]p

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Example

R=











f(g(x),g(y),a) 7→_r1 j(x,x,a)

g(b) 7→_r2 b

h(b) 7→_r3 b

The critical pairs are in red:

j(b,b,a)r1← f(g(b),g(y),a)→_r2 f(b,g(y),a) (r1,r2,pos=1) j(x,x,a)r1← f(g(x),g(b),a)→_r2 f(g(x),b,a) (r1,r2,pos=2)

Putting the information in a table:

r1 r2 r3

r1 X positions=1,2 X

r2 X X

r3 X

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Example

R=











f(f(x)) 7→_r1 g(x) f(b) 7→_r2 c

b 7→_r3 a

The critical pairs are in red:

g(f(x)) r1← f(f(f(x))) →_r1 f(g(x)) (r1,r1,pos=1) g(b) r1← f(f(b)) →_r2 f(c) (r1,r2,pos=1) c r2← f(b) →_r3 f(a) (r2,r3,pos=1)

Putting the information in a table:

r1 r2 r3

r1 positions=1 positions=1 X

r2 X positions=1

r3 X

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Local confluence by critical pairs

Theorem

LetRbe a rewrite system. ThenRis locally confluent iff every critical pair ofRis joinable.

Proof.

Theonly ifimplication is trivial.

For theifimplication, let us take any case of the form

v←t→ u Three cases are possible:

Disjoint reductions:

f(b,h(b),a) ← f(g(b),h(b),a) → f(g(b),b,a) f(b,h(b),a) → f(b,b,a) ← f(g(b),b,a)

Not disjoint and not critical:

j(h(b),h(b),a) ← f(g(h(b)),g(b),a) → f(g(b),g(b),a)

↓ ↓

j(b,h(b),a) → j(b,b,a)

Not disjoint and critical: we close the diagram by the hypothesis.

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Decidable case of confluence

Theorem

LetRbe afiniteandSNrewrite system. Then confluence ofRis decidable.

Proof.

The algorithm:

1 Generate all the critical pairs.

2 For each critical pairhu,vi, if there issomenormal formbuofuandsomenormal formbvofvsuch thatbu,bv, thenfail.

3 Otherwise (no fail for some critical pair),succeed.

Remark that

If the algorithm fails, then there is a critical pair which is not joinable, soRis not confluent by the previous theorem.

If the algorithm succeeds, then every critical pair is joinable, so thatRis locally confluent by the previous theorem. To obtain confluence, apply Newmann’s Lemma.

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Orthogonal Systems

A system isorthogonaliff it isleft linear(no duplication of variables on the left of rules) andhas no critical pairs.

Example

0+y 7→ y

s(x)+y 7→ s(x+y)

0∗y 7→ 0

s(x)∗y 7→ (x∗y)+y

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Confluence by orthogonality

Theorem

IfRis orthogonal, then it is confluent.

Proof.

We define a notion ofparallel reductionassociated toRfor first-order terms:

s_Rs (R) l7→r∈ Randσa subst.

σ(l)_Rσ(r) (S) s1Rt1. . . .snRtn

f(s1, . . . ,sn)_R f(t1, . . . ,tn) (C) Example:

ForR={f(x,y)7→h(y,y),a7→b}, we have f(f(a,c),a,f(a,c))R f(h(c,c),b,f(b,c))

Now use theconfluence by equivalencetechnique:

1 Observe that→R ⊆ R ⊆ →^∗_R.

2 Show thatRhas the diamond property.

3 Conclude that→R is confluent.

Obs: from now on, we simply writeinsted ofRisRis clear from the context.

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The relation has the diamond property

Defineσδfor two substitutions iffdom(σ)=dom(δ)andσxδxfor everyx∈dom(σ).

Lemma (A)

Ifσδ, thenσtδtfor every termt.

Proof.

By induction ont.

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Lemma (B)

LetRbe an orthogonal rewriting system. Letσbe a substitution. Letsbe a strict subterm ofl, wherel7→r∈ R. Ifσst, then there isδs.t.t=δsandσδ.

The proof is by induction ons.

Ifs=x, then defineδx=tandδy=σyfory,x.

Ifs= f(s1, . . . ,sn), we distinguish three cases according to the definition ofσst. Ifσstholds by(R), thent=σsand thus we simply letδ:=σ.

Ifσstholds by(S), then there is some ruleg7→d∈ Rand some substitutionτsuch thatσs=τgandt=τd. We can assumel7→randg7→ddo not share variables and thus(σ∪τ)s=σs=τg=(σ∪τ)g. Thenσ∪τunifiess(a strict subterm ofl) andg, thus contradictingabsence of critical pairs, and thusorthogonalityofR.

Ifσstholds by(C), thent=f(t1, . . . ,t_n)andσsit_i. Sincel(and thuss) arelinear, we can writeσasSσi, where eachσiisσrestricted to the variables ofsi. We then haveσis_it_iands_iis a strict subterm ofl. The i.h. givest_i=δis_iandσiδi. We take δ:=Sδi. It is easy to checkσδandt=δs.

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Theorem

LetRbe an orthogonal system andits associated parallel reduction relation. The reduction relationhas the diamond property.

Proof.

Suposett1andtt2. We reason by cases.

If one of them uses(R), then we trivially close the diagram.

If both use(S), thent=σl,t1=σr,t=δgandt2=δdforl7→randg7→dinR. If the two rules are the same, thenσ=δand we trivially close the diagram.

Otherwise, we can assume that the rules do not share variables so

that(δ∪σ)l=σl=t=δg=(σ∪δ)g,i.e.(δ∪σ)unifieslandg, which contradicts absence of critical pairs, and thusorthogonalityofR_.

If both use(C), the property trivially holds by the i.h.

Suppose one uses(C)and the other one uses(S). We have

t=f(v1, . . . ,vn) f(u1, . . . ,un)=t1whereviuiandt=σlσr=t2for some rulel7→r∈ Rand some substitutionσ. The left-hand sidelis necessarily of the form f(l1, . . . ,ln)so thatt=f(v1, . . . ,vn)=f(σl1, . . . , σln)andσliui. Butlis linearso that we can writeσ=Sn

1σiwhereσiliui. Each termliis a strict subterm oflso that byLemma (B)there are substitutionsδisuch thatui=δiliand σiδi. We letδ:=Sn

1δi. Thent1=f(δ1l1, . . . , δnln)=δland alsoσδ. We can now close the diagram witht3=δras follows

t₁=δlδr=t₃holds using the(S)rule.

t₂=σrδr=t₃holds byLemma (A)using the fact thatσδ.

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Important remark (I)

Left linearity alone is not sufficient for confluence.

Example

a 7→ b b 7→ a c 7→ a

a 7→ c b 7→ d c 7→ e

Two not joinable terms:

e ^∗←a→^∗ d

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Important remark (II)

Absence of critical pairs is not sufficient for confluence.

Example

f(x,x) 7→ a f(x,g(x)) 7→ b

c 7→ g(c)

Two not joinable terms:

b ^∗← f(c,c)→^∗a

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Relaxing orthogonality

por(t,x) 7→ t por(x,t) 7→ t This system is not orthogonal but the critical pair is trivial.

A systemRisparallel closediff for every critical pairhu,viofRwe havevu.

Theorem

IfRis left-linear and parallel closed then it is confluent.

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Confluence by decreasing diagrams

Vincent van Oostrom (1994) Idea: every reduction step islabelled

wheret→_gS umeanst→_s ufor somes<S,i.e.∃i∈ Ssuch thats<i. As a consequence,t→^∗

glumeans that every step fromttouis labelled with somes⁰<l, andt→^∗_g_{l,m}umeans that every step fromttouis labelled with somes⁰<max(l,m)(ifl andmare comparable).