Termination w.r.t. Q proof of /tmp/tmp1oj7JS/Ex1_Luc04b

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

0 QTRS

↳1 QTRSRRRProof (⇔)

↳2 QTRS

↳3 QTRSRRRProof (⇔)

↳4 QTRS

↳5 QTRSRRRProof (⇔)

↳6 QTRS

↳7 QTRSRRRProof (⇔)

↳8 QTRS

↳9 DependencyPairsProof (⇔)

↳10 QDP

↳11 DependencyGraphProof (⇔)

↳12 QDP

↳13 MRRProof (⇔)

↳14 QDP

↳15 DependencyGraphProof (⇔)

↳16 QDP

↳17 QDPOrderProof (⇔)

↳18 QDP

↳19 QDPOrderProof (⇔)

↳20 QDP

↳21 QDPOrderProof (⇔)

↳22 QDP

↳23 DependencyGraphProof (⇔)

↳24 TRUE

(0) Obligation:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__head(x₁)) = 1 + x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(a__tail(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(head(x₁)) = 1 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)

(2) Obligation:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__head(x₁)) = 2 + x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(a__tail(x₁)) = 2 + 2·x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(head(x₁)) = 2 + x₁
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁
POL(nats) = 0
POL(nil) = 2
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + 2·x₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(head(X)) → a__head(mark(X))
mark(nil) → nil
a__tail(X) → tail(X)

(4) Obligation:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__head(x₁)) = 2 + 2·x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 0
POL(a__odds) = 0
POL(a__pairs) = 0
POL(a__tail(x₁)) = 1 + 2·x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(head(x₁)) = 1 + x₁
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2 + 2·x₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(tail(X)) → a__tail(mark(X))
a__head(X) → head(X)

(6) Obligation:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__incr(x₁)) = 2·x₁
POL(a__nats) = 2
POL(a__odds) = 0
POL(a__pairs) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁
POL(nats) = 1
POL(odds) = 0
POL(pairs) = 0
POL(s(x₁)) = x₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__nats → nats

(8) Obligation:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

A__ODDS → A__INCR(a__pairs)
A__ODDS → A__PAIRS
A__INCR(cons(X, XS)) → MARK(X)
MARK(nats) → A__NATS
MARK(pairs) → A__PAIRS
MARK(odds) → A__ODDS
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(odds) → A__ODDS
A__ODDS → A__INCR(a__pairs)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(13) 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 dependency pairs:

A__ODDS → A__INCR(a__pairs)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(A__ODDS) = 2
POL(MARK(x₁)) = 2·x₁
POL(a__incr(x₁)) = x₁
POL(a__nats) = 0
POL(a__odds) = 1
POL(a__pairs) = 1
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(odds) = 1
POL(pairs) = 1
POL(s(x₁)) = x₁

(14) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(odds) → A__ODDS
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 1 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(A__INCR(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(s(x₁)) =
/ 1 \

\ 0 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(a__incr(x₁)) =
/ 1 \

\ 1 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(a__odds) =
/ 1 \

\ 1 /

POL(odds) =
/ 1 \

\ 1 /

POL(a__pairs) =
/ 0 \

\ 0 /

POL(pairs) =
/ 0 \

\ 0 /

POL(0) =
/ 0 \

\ 0 /

POL(a__nats) =
/ 0 \

\ 0 /

POL(nats) =
/ 0 \

\ 0 /

The following usable rules [FROCOS05] were oriented:

a__incr(X) → incr(X)
a__odds → odds
a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(pairs) → a__pairs
a__pairs → cons(0, incr(odds))
a__nats → cons(0, incr(nats))
mark(nats) → a__nats
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__odds → a__incr(a__pairs)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(incr(X)) → a__incr(mark(X))
mark(odds) → a__odds

(18) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(A__INCR(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 1 0 /

· x₁

POL(cons(x₁, x₂)) =
/ 1 \

\ 0 /

+
/ 1 0 \

\ 1 0 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(a__incr(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(a__odds) =
/ 1 \

\ 0 /

POL(odds) =
/ 1 \

\ 0 /

POL(a__pairs) =
/ 1 \

\ 0 /

POL(pairs) =
/ 1 \

\ 0 /

POL(0) =
/ 0 \

\ 0 /

POL(a__nats) =
/ 1 \

\ 0 /

POL(nats) =
/ 1 \

\ 0 /

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

The following usable rules [FROCOS05] were oriented:

a__incr(X) → incr(X)
a__odds → odds
a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(pairs) → a__pairs
a__pairs → cons(0, incr(odds))
a__nats → cons(0, incr(nats))
mark(nats) → a__nats
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__odds → a__incr(a__pairs)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(incr(X)) → a__incr(mark(X))
mark(odds) → a__odds

(20) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 1 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(A__INCR(x₁)) =
/ 1 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(a__incr(x₁)) =
/ 0 \

\ 1 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(a__odds) =
/ 0 \

\ 1 /

POL(odds) =
/ 0 \

\ 1 /

POL(a__pairs) =
/ 0 \

\ 0 /

POL(pairs) =
/ 0 \

\ 0 /

POL(0) =
/ 0 \

\ 0 /

POL(a__nats) =
/ 0 \

\ 0 /

POL(nats) =
/ 1 \

\ 1 /

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁

The following usable rules [FROCOS05] were oriented:

a__incr(X) → incr(X)
a__odds → odds
a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(pairs) → a__pairs
a__pairs → cons(0, incr(odds))
a__nats → cons(0, incr(nats))
mark(nats) → a__nats
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__odds → a__incr(a__pairs)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(incr(X)) → a__incr(mark(X))
mark(odds) → a__odds

(22) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) QTRSRRRProof (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) QTRSRRRProof (EQUIVALENT transformation)

(6) Obligation:

(7) QTRSRRRProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyPairsProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) MRRProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) DependencyGraphProof (EQUIVALENT transformation)

(24) TRUE