Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
      ↳26 Instantiation (⇔)
        ↳27 QDP
      ↳28 Instantiation (⇔)
        ↳29 QDP
        ↳30 Instantiation (⇔)
        ↳31 QDP
        ↳32 NonTerminationProof (⇔)
        ↳33 FALSE
  ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 QReductionProof (⇔)
    ↳38 QDP
    ↳39 QDPSizeChangeProof (⇔)
    ↳40 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FST(s(X), cons(Y, Z)) → FST(X, Z)
FROM(X) → FROM(s(X))
ADD(s(X), Y) → ADD(X, Y)
LEN(cons(X, Z)) → LEN(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LEN(cons(X, Z)) → LEN(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LEN(cons(X, Z)) → LEN(Z)

R is empty.
The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LEN(cons(X, Z)) → LEN(Z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LEN(cons(X, Z)) → LEN(Z)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

R is empty.
The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ADD(s(X), Y) → ADD(X, Y)
    The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(X) → FROM(s(X))

R is empty.
The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(X) → FROM(s(X))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FROM(X) → FROM(s(X)) we obtained the following new rules [LPAR04]:

FROM(s(z0)) → FROM(s(s(z0)))

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(s(z0)) → FROM(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FROM(X) → FROM(s(X)) we obtained the following new rules [LPAR04]:

FROM(s(z0)) → FROM(s(s(z0)))

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(s(z0)) → FROM(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(30) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FROM(s(z0)) → FROM(s(s(z0))) we obtained the following new rules [LPAR04]:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

(31) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(32) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = FROM(s(s(z0))) evaluates to t =FROM(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [z0 / s(z0)]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FROM(s(s(z0))) to FROM(s(s(s(z0)))).



(33) FALSE

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FST(s(X), cons(Y, Z)) → FST(X, Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(35) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FST(s(X), cons(Y, Z)) → FST(X, Z)

R is empty.
The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(37) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

(38) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FST(s(X), cons(Y, Z)) → FST(X, Z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(39) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • FST(s(X), cons(Y, Z)) → FST(X, Z)
    The graph contains the following edges 1 > 1, 2 > 2

(40) TRUE