Term Rewriting System R:
[X, Y, Z]
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
DBL(s(X)) -> DBL(X)
DBLS(cons(X, Y)) -> DBL(X)
DBLS(cons(X, Y)) -> DBLS(Y)
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
INDX(cons(X, Y), Z) -> SEL(X, Z)
INDX(cons(X, Y), Z) -> INDX(Y, Z)
FROM(X) -> FROM(s(X))
DBL1(s(X)) -> DBL1(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, Z)
QUOTE(s(X)) -> QUOTE(X)
QUOTE(dbl(X)) -> DBL1(X)
QUOTE(sel(X, Y)) -> SEL1(X, Y)
Furthermore, R contains eight SCCs.
R
↳DPs
→DP Problem 1
↳Size-Change Principle
→DP Problem 2
↳SCP
→DP Problem 3
↳MRR
Dependency Pair:
DBL(s(X)) -> DBL(X)
Rules:
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
We number the DPs as follows:
- DBL(s(X)) -> DBL(X)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
D_{P}: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
s(x_{1}) -> s(x_{1})
We obtain no new DP problems.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳Size-Change Principle
→DP Problem 3
↳MRR
Dependency Pair:
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
Rules:
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
We number the DPs as follows:
- SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
D_{P}: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
cons(x_{1}, x_{2}) -> cons(x_{1}, x_{2})
s(x_{1}) -> s(x_{1})
We obtain no new DP problems.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳SCP
→DP Problem 3
↳Modular Removal of Rules
Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
We have the following set of usable rules:
none
To remove rules and DPs from this DP problem we used the following monotonic and C_{E}-compatible order: Polynomial ordering.
Polynomial interpretation:
_{ }^{ }POL(FROM(x_{1})) | = x_{1}_{ }^{ } |
_{ }^{ }POL(s(x_{1})) | = x_{1}_{ }^{ } |
We have the following set D of usable symbols: {FROM, s}
No Dependency Pairs can be deleted.
17 non usable rules have been deleted.
The result of this processor delivers one new DP problem.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳SCP
→DP Problem 3
↳MRR
→DP Problem 9
↳Non-Overlappingness Check
Dependency Pair:
FROM(X) -> FROM(s(X))
Rule:
none
R does not overlap into P. Moreover, R is locally confluent (all critical pairs are trivially joinable).Hence we can switch to innermost.
The transformation is resulting in one subcycle:
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳SCP
→DP Problem 3
↳MRR
→DP Problem 9
↳NOC
...
→DP Problem 10
↳Non Termination
Dependency Pair:
FROM(X) -> FROM(s(X))
Rule:
none
Strategy:
innermost
Found an infinite P-chain over R:
P =
FROM(X) -> FROM(s(X))
R = none
s = FROM(X)
evaluates to t =FROM(s(X))
Thus, s starts an infinite chain as s matches t.
Non-Termination of R could be shown.
Duration:
0:03 minutes