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 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔)
        ↳16 QDP
        ↳17 QReductionProof (⇔)
        ↳18 QDP
        ↳19 UsableRulesReductionPairsProof (⇔)
        ↳20 QDP
        ↳21 QReductionProof (⇔)
        ↳22 QDP
        ↳23 MNOCProof (⇔)
        ↳24 QDP
        ↳25 NonTerminationProof (⇔)
        ↳26 NO

(0) Obligation:

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

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

LEQ(s(X), s(Y)) → LEQ(X, Y)
DIFF(X, Y) → IF(leq(X, Y), 0, s(diff(p(X), Y)))
DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → DIFF(p(X), Y)
DIFF(X, Y) → P(X)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LEQ(s(X), s(Y)) → LEQ(X, Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

LEQ(s(X), s(Y)) → LEQ(X, Y)

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

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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].

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(x0, x1)

(11) Obligation:

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

LEQ(s(X), s(Y)) → LEQ(X, Y)

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:

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

(13) YES

(14) Obligation:

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

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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].

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(x0, x1)

(18) Obligation:

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

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(19) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

p(s(X)) → X
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIFF(x1, x2)) = 2·x1 + x2   
POL(p(x1)) = x1   
POL(s(x1)) = x1   

(20) Obligation:

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

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.[THIEMANN].

p(s(x0))

(22) Obligation:

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

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0

The set Q consists of the following terms:

p(0)

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

(23) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(24) Obligation:

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

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0

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

(25) 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 = DIFF(X, Y) evaluates to t =DIFF(p(X), Y)

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from DIFF(X, Y) to DIFF(p(X), Y).



(26) NO