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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 AND
  ↳9 QDP
    ↳10 UsableRulesProof (⇔)
    ↳11 QDP
    ↳12 QReductionProof (⇔)
    ↳13 QDP
    ↳14 NonTerminationProof (⇔)
    ↳15 FALSE
  ↳16 QDP
    ↳17 UsableRulesProof (⇔)
    ↳18 QDP
    ↳19 QReductionProof (⇔)
    ↳20 QDP
    ↳21 QDPSizeChangeProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 UsableRulesProof (⇔)
    ↳25 QDP
    ↳26 QReductionProof (⇔)
    ↳27 QDP
    ↳28 QDPSizeChangeProof (⇔)
    ↳29 TRUE

(0) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(adx(x1)) = 1 + x1   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(nats) = 2   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

adx(nil) → nil
nats → adx(zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L


(2) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

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

(4) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

INCR(cons(X, L)) → INCR(L)
ADX(cons(X, L)) → INCR(cons(X, adx(L)))
ADX(cons(X, L)) → ADX(L)
ZEROS → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node.

(8) Complex Obligation (AND)

(9) Obligation:

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

ZEROS → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(10) 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.

(11) Obligation:

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

ZEROS → ZEROS

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

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(12) 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].

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

(13) Obligation:

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

ZEROS → ZEROS

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

(14) 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 = ZEROS evaluates to t =ZEROS

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ZEROS to ZEROS.



(15) FALSE

(16) Obligation:

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

INCR(cons(X, L)) → INCR(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(17) 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.

(18) Obligation:

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

INCR(cons(X, L)) → INCR(L)

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

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(19) 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].

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

(20) Obligation:

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

INCR(cons(X, L)) → INCR(L)

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

(21) 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:

  • INCR(cons(X, L)) → INCR(L)
    The graph contains the following edges 1 > 1

(22) TRUE

(23) Obligation:

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

ADX(cons(X, L)) → ADX(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(cons(X, L)) → incr(cons(X, adx(L)))
zeros → cons(0, zeros)

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(24) 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.

(25) Obligation:

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

ADX(cons(X, L)) → ADX(L)

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

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

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

(26) 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].

incr(nil)
incr(cons(x0, x1))
adx(cons(x0, x1))
zeros

(27) Obligation:

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

ADX(cons(X, L)) → ADX(L)

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

(28) 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:

  • ADX(cons(X, L)) → ADX(L)
    The graph contains the following edges 1 > 1

(29) TRUE