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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 ForwardInstantiation (⇔)
    ↳12 QDP
    ↳13 UsableRulesProof (⇔)
    ↳14 QDP
    ↳15 MRRProof (⇔)
    ↳16 QDP
    ↳17 NonTerminationProof (⇔)
    ↳18 FALSE

(0) Obligation:

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

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

F(s(0), g(x)) → F(x, g(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

G(s(x)) → G(x)

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

(8) 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:

  • G(s(x)) → G(x)
    The graph contains the following edges 1 > 1

(9) TRUE

(10) Obligation:

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

F(s(0), g(x)) → F(x, g(x))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

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

(11) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(s(0), g(x)) → F(x, g(x)) we obtained the following new rules [LPAR04]:

F(s(0), g(s(0))) → F(s(0), g(s(0)))

(12) Obligation:

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

F(s(0), g(s(0))) → F(s(0), g(s(0)))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

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

(13) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

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

F(s(0), g(s(0))) → F(s(0), g(s(0)))

The TRS R consists of the following rules:

g(s(x)) → g(x)

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

(15) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

g(s(x)) → g(x)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2)) = x1 + x2   
POL(g(x1)) = x1   
POL(s(x1)) = 1 + x1   

(16) Obligation:

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

F(s(0), g(s(0))) → F(s(0), g(s(0)))

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

(17) 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 = F(s(0), g(s(0))) evaluates to t =F(s(0), g(s(0)))

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(s(0), g(s(0))) to F(s(0), g(s(0))).



(18) FALSE