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jan willem klop

roel de vrijer

infinitarynormalization

TeR, 1 y 8 junio 2007

Finite and infinite reductions

TRS: f(x,y) f(y,x)

f(a,b) f(b, a) f(a,b) f(b, a) …

f(a,a) f(a, a) f(a,a) f(a, a) …

TRS: c g(c)

c g(c) g(g(c)) g(g(g(c)))) …

3

Cauchy converging reduction sequence: activity may occur everywhere

Strongly converging reduction sequence, with descendant relations

Transfinite reductions

TRS: c g(c)

c g(c) g(g(c)) g(g(g(c)))) … g

The last “step” is a limit transition

We have:

c g

gg

gg

5

F(0) P

0 F

S

0

P

P

P

P

0

S

0 S

S

0

.....

Limit: infinite sequence of natural numbers

TRS: F(x) → P(x, F(S(x)))

Finite and infinite terms

term = term tree = set of labeled positions

1 the set of positions is closed under prefixes2 each position is labeled by a function

symbol or a variable3 the arity of the function symbol at a position

equals the number of outgoing edges

Streams

x : y infix notation for P(x,y)

F(0) → 0 : F(S(0)) → 0 : S(0) : F(S(S(0))) → 0 : S(0) : S(S(0)) : F(S(S(S(0)))) → ….

→ω 0 : S(0) : S(S(0)) : S(S(S(0))) : ….

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zip(zeros, ones) and alt rewrite in infinitely many steps to the same infinite normal form

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P

F F

0 0

P

F

0

PP

PP

P

0S0 S

S0

PP

PP

P

0S0 S

S0

P

PP

PP

0S0 S

S0

P

Transfinite reduction sequence of length

Transfinite reductions

TRS: c g(c) c g

c : c : c : ….

g : c : c : ….

g : g : c : ….

g : g : g : …. …….

c : c : c : c : … . g : g : g : g : …

Convergence and divergence

We have ρ: t t at limit ordinal if:

1. For each limit ordinal λ < β the prefix βλ is convergent

2. (dα)α < tends to infinity

3. the limit is t

ρ is divergent if 1. but not 2.

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0 ·2 ·3 2

depth of contracted redex tends to infinity at each limit ordinal

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Every countable ordinal can be the length of a transfinite reduction.

TRS: c f(a, c) a b

f

faa f

faaBut no uncountable

reductions!-- only finitely many reduction steps at depth n-- no infinite descending chains of ordinals

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For orthogonal TRSs we have the

Compression Lemma:

every transfinite reduction of length can be compressed to or less.

Left-linearity is essential for compression:

TRS: f(x,x) c a g(a) b g(b)

f(a,b) f(g,g) c

So f(a,b) +1 c

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How to define SN and WN?

WN is easy: There is a possibly infinite reduction to a possibly infinite normal form.

SN: all reductions will eventually terminate in a (infinitary) normal form ?

?

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Examples of SN

a g(a)

c f(a, c) and a b

S-terms in CL

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Example of not SN

I(x) x

This TRS is SN, but:

I I I I …

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Good and bad reductions. In ordinary rewriting the finite reductions are good, they have an end point. The infinite ones are bad, they have no end point.

In infinitary rewriting the good reductions are the ones that are strongly convergent, they have an end point:

a b(a) reaches after steps the end point b.

The bad reductions (divergent) are the ones without end point. These reductions may be long, a limit ordinal long, but there they fail.

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What is a divergent (bad) reduction?

- Select step after step a redex and perform it

- Go on until a limit ordinal - At that point look back

- If strongly convergent: take the limit and go on

- If not, stop there: we found a divergent reduction

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Divergent reduction (again)

Suppose you have for each ordinal < limit ordinal (convergent) reductions that extend each other

Suppose there is no limit for these reductions

Together these reductions constitute a divergent reduction of length

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SN: there are no divergent reductions

So a divergent reduction is counterexample to SN

(such as an infinite reduction is counterexample to SN)

Important note:Given a divergent reduction we can find a term where infinitely often a root step was performed.

Divergence: (dα)α < does not tend to infinity

n. <. >. d n, take N smallest such n and take s.t. >. d ≥ N, then from :

1. Infinitely many steps at depth N2. Finitely many positions at depth N3. At one of these positions infinitely many root steps (head steps)

Conclusion: a divergent reduction with infinitely many head steps

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Failure of Newman’s Lemma, infinitary version

We do not have the implication WCR & SN CR .

I.e. Newman’s Lemma, WCR & SN CR, fails for infinitary term rewriting.

C A(C)A(C) B(C)A(B(x)) B(A(x))

Note that we do not have A(x) B(x)!

Check WCR by looking at the critical pairs. We also have SN.

But CR∞ fails, as C reduces in steps to A and B

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A(C)

A(A(C))B(C)

C A(C)A(C) B(C)A(B(x)) B(A(x)).

A(B(C))

B(A(C))

Huet’s critical pair lemma

WCR

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A(x) x B(x) x C A(B(C))

C A(B(C))

A(C) B(C) A(A(B(C))) B(A(B(C))) A(A(C)) B(B(C)) A(A(A(B(C)))) B(B(A(B(C)))) A(A(A(C))) B(B(B(C))) A B

......

......

C ABC ABABC ABABABC ABABABABAB...

A B

...

(a) (b)

Failure of infinitary confluence

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Sxyz xz(yz) Kxy x

@(@(@(S, x), y), z) @(@(x, z), @(y, z)) @(@(K, x), y) x

@

@ K

K

@

@ S

K

@

@ K

K @

@ S

K @

@ K

K @

@ S

K @

@

@ S

K @

@ S

K @

@ S

K @

@ S

K @

@

@ K

K @

@ K

K @

@ K

K @

@ K

K @

collapsing contexts

Failure of infinitary confluence for Combinatory Logic

Non-confluent infinite reduction diagram

for OTRSs: PML

Projection of convergent reduction against parallel step

Projection of a parallel step over infinite reduction

can result in infinite development:

TRS: f(x) x : f(x) a b

f(a) a : a : a : a … 1. Redex a has infinitely many residuals2. But all are parallel !!3. Development of a parallel set of redexes is always convergent

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Developments are not always convergent

I(x) x

Consider in I the set of all redexes

I I I I …

for OTRSs: UN

A term t has at most 1 infinite normal form

Easy consequence:

SN => CR

Proof of UN

C-stable reduction: All activity below prefix CC-stable term t: From t only C-stable reductionn-stable: Stable for full prefix of depth n

Lemma: if t has a C-stable reduction to nf, then t is C-stable

Redex overlapping with prefix

Proof of Lemma: t has C-stable reduction to nf => t is C-stable

1. No redex in t overlaps with C2. Projection of C-stable over C-stable is C-stable3. If t → s then - C is prefix of s - no redex of s overlaps with C

4. Repeating this: t is C-stable

Note that we used PML

Proof of UN

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THEOREM. SN WN

is clear

Suppose not SN

So there is a term with an infinite reduction in which infinitely many root steps occur

Such a term does not have an (infinite) normal form,

i.e. not WN. This uses the Head Normalization Theorem generalized to infinite terms

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Head Normalization Theorem

THEOREM. In an orthogonal TRS let t t’ t” ...

be a reduction containing infinitely many head steps.

(i) Then t has no head normal form.(ii) Consequence: t has no (infinitary) normal form.

t is a head normal form (hnf) when it does not reduce to a redex; i.e. no reduction from t contains a root (head) step

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Head Normalization Theorem: proof

- Reduction to head normal form can always be finite

- Projection of

t0 t1 t2 ... with infinitely many head steps over parallel step yields infinitely many head steps again

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Projection of head step over parellel step t t+1

↓ p ↓ p+1

t’ t+1’

Either of two possibilities:1. p internal, then projection of head step again head step2. p head step, then p+1 empty step

Projecting ρ with infinitely many head steps:- If 2. occurs in ρ, then the projection is eventually ρ- If 2. never occurs, then in the projection infinitely many head steps by 1.