Basic definitions

Chapter 2

Basic definitions

2.1 Terms and substitutions 2.2 Term Rewriting systems

A term rewriting systemRis a set of pairs of terms inT(F, X). Its members are typically written l→ r

The following relations are defined onT(F, X).

• s −−→^p,σ

l→r tif s|p=lσ and t=s[rσ]p

• s −−→

l→r t if there is a position p ∈ Pos(s) and a substitution σ such that s −−→^p,σ

l→r t.

• s −→

R tif there is a rule l→r∈Rsuch thats −−→

l→r t

• ←−−→

R = −→

R ∪ ←−

R

• −→^∗

R =!+∞

n=0

−→n

R where −→⁰

R is the identity and −−−→ⁿ⁺¹

R = −→

R ◦ −→ⁿ

R . Example 1

(r₁) dec(enc(x, y), y) → x (r₂) π₁('x, y() → x (r3) π₂('x, y() → y dec(enc(π₁('a, b(), a), a) ",{x$→π1(%a,b&);y$→a}

−−−−−−−−−−−−−→

r1

π₁('a, b() 5

6 CHAPTER 2. BASIC DEFINITIONS Definition 1 A term rewriting system R is terminating if there is no infi- nite sequence{si}i∈N such that, for every i, si −→

R s_i+1.

Note that this definition corresponds touniversall terminationandstrong normalization; we may start from an arbitrary term and the reductions take place at any position, using any rule.

Exercice 1

Give an example of a finite TRSR, which is not terminating and such that, for every termt, there is a termusuch thatt −→^∗

R uanducannot be reduced byR.

Exercice 2

Give an example of a finite TRS which 1. is not terminating

2. each rule alone is a terminating system

3. for any term tand any position poft, at most one rule can be applied at position pin t

Theorem 1 Termination is undecidable for finite term rewriting systems.

Proof:

We reduce the Post Correspondance Problem. Let (u1, . . . , u_n),(v1, . . . , v_n) be an instance of PCP, whereu_i, v_i ∈Σ^∗.

We consider the set of symbols F = {0(0), f(4)}∪{a(1)|a ∈ Σ}. If u ∈ Σ^∗ and t ∈ T(F, X), we write u(t) the term defined by induction on u: #(t) = t, au(t) = a(u(t)). u(t) is defined by induction on" u: "#(t) = t,

#

au(t) =u(a(t))."

We let Rbe the rewrite system containing the rules

$ (r₁ⁱ) f(u"i(x),v"i(y), x₁, y₁) → f(x, y, ui(x₁), vi(y₁)) For every pair (ui, vi) (r₂^a) f(x, y, a(z), a(z)) → f(a(x), a(y), z, z) For every letter a We claim that PCP has a solution iffRis not terminating.

If PCP has a solution u_i1· · ·u_ik =v_i1· · ·v_ik =w, then f(u_i!₁· · ·u_ik(0),v_i1!· · ·v_ik(0),0,0) −−→

r^ik₁

· · · −−→

rⁱ₁¹

f(0,0, u_i1· · ·u_ik(0), v_i1· · ·v_ik(0))

2.2. TERM REWRITING SYSTEMS 7 and

f(0,0, w(0), w(0)) −→^∗

r2

f(w(0)," w(0)," 0,0) =f(u_i!₁· · ·u_ik(0),v_i!₁· · ·v_ik(0),0,0) Hence Ris not terminating.

If R is not terminating , consider a term tfrom which there is an infi- nite reduction sequence.

t −−−→^p1,σ1

R t₁· · · −−−→^pn,σn

R t_n· · ·

We first note that the number of occurrences off inti(written #_f(ti)) is constant along the sequence since the reqwriting rules in R do not erase nor duplicate variables:

#_f(u[lσ]p) = #_f(u[0]p) + #_f(l) +%

x∈V ar(l)#_f(xσ)

= #_f(u[0]p) + #_f(r) +%

x∈V ar(r)#_f(xσ)

= #_f(u[rσ]p)

Now, we prove, by induction on (#_f(t),|t|), (where |t| is the size of the term t) that there is an infinite reduction sequence

s₁ −→^" s₂· · · −→^" s_n −→^"

in which all reductions take place at the root position.

If #_f(t) = 0, there is nothing to prove.

If t=a(t⁽) for somea∈Σ, thenp_i = 1·p⁽_i for all iand t|1

p^!1,σ1

−−−→R · · ·t_n|1 p^!_n,σn

−−−→R · · ·

and we can apply the induction hypothesis tot|1, whose size is strictly smaller than the size of t.

If t = f(α,β,γ,δ). Then, for every i, t_i = f(α_i,β_i,γ_i,δ_i). Consider again two cases: either {i∈N|p_i =#} is infinite or not.

Case 1: {i∈N|pi =#} is finite . Let i₀ be the maximum of this set. Then, for i > i₀, p_i > #. Therefore, one of the four sets {pi|i > i₀}∩j.N^∗ forj = 1,2,3,4 is infinite: we can extract an infinite sequence

t_m1|j −→

R · · ·t_mp|j −→

R · · ·

and #_f(t_m1|j)<#f(t); we can apply the induction hypothesis.

8 CHAPTER 2. BASIC DEFINITIONS Case 2: {i∈N|p_i =#} is infinite . Consider the morphismρ such that, ρ(a(t)) = a(ρ(t)) for a ∈ Σ and ρ(f(s₁, . . . , s₄)) = 0.

Thanks to the definition of R, ifs −→

R s⁽, then ρ(s) =ρ(s⁽).

We let ρ⁽ be the mapping defined by ρ⁽(a(t)) = ρ(a(t)) and ρ⁽(f(s₁, s₂, s₃, s₄)) = f(ρ(s₁),ρ(s₂),ρ(s₃),ρ(s₄)). Let us show that, for everyi, eithert_i −→^" t_i+1, in which caseρ⁽(ti) −→^" ρ⁽(ti+1) or else ρ⁽(t_i) =ρ⁽(t_i+1).

Indeed, by definition of R, ρ⁽(lσ) = lσ^ρ where xσ^ρ = ρ(xσ) for every variable x. Hence, if ti =lσi, then ρ⁽(ti) =lσ^ρ_i −→^" rσ_i^ρ = t_i+1. Ifti

)="

−−→ti+1

, then, forj= 1,2,3,4,ρ(ti|j) =ρ(t_i+1|j), hence ρ⁽(ti) =ρ⁽(t_i+1).

Therefore, ifti_kis the subsequence of terms such thatti_k "

−→R ti_k+1, then

ρ⁽(ti1) −→^"

R ρ⁽(ti2) −→^"

R · · · −→^"

R ρ⁽(ti_k) −→^"

R · · · We are left now to consider the case where all p_i=#.

Consider the two interpretations: I₁(t_i) = (|αi|,|βi|,|γi|,|δi|)lex and I₂(ti) = (|δi|,|γi|,|βi|,|αi|)lex. The lexicographic ordering on N⁴ is well-founded and, if s −→^"

rⁱ₁

s⁽, then I₁(s)> I₁(s⁽) and, if −→^"

r₂^a s⁽, then I₂(s) > I₂(s⁽). It follows that there is no infinite reduction sequence using the rulesrⁱ₁only, nor using the rulesr₂^aonly. In other words, the infinite reduction sequence must swicth infinitely often between ther₁ rules and the r₂ rules. Therefore, there is a subsequence

f(u, v, a(w), a(w)) −→

r2

f(a(u), a(v), w, w)

"

−−→

r₁^ik

f(u⁽₁, v₁⁽, u_ik(w), vik(w))

· · ·

"

−−→

r₁ⁱ¹

f(u⁽_k, v⁽_k, u_i1· · ·u_ik(w), v_i1· · ·v_ik(w))

"

−→r^a₂ · · ·

But applying a ruler₂^aat position#tof(u⁽_k, v_k⁽, u_i1· · ·u_ik(w), vi1· · ·v_ik(w)) requiresu_i1· · ·u_ik(w) =v_i1· · ·v_ik(w), which impliesu_i1· · ·u_ik =v_i1· · ·v_ik: there is a solution tp PCP.