YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex15_Luc98_GM.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 67 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 0 ms)
↳6 QTRS
↳7 QTRSRRRProof (⇔, 0 ms)
↳8 QTRS
↳9 QTRSRRRProof (⇔, 0 ms)
↳10 QTRS
↳11 RisEmptyProof (⇔, 0 ms)
↳12 YES

(0) Obligation:

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

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(a__add(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(a__and(x1, x2)) = 2 + x1 + 2·x2   
POL(a__first(x1, x2)) = 2 + x1 + x2   
POL(a__from(x1)) = 2 + 2·x1   
POL(a__if(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3   
POL(add(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(and(x1, x2)) = 2 + x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(false) = 2   
POL(first(x1, x2)) = 1 + x1 + x2   
POL(from(x1)) = 1 + x1   
POL(if(x1, x2, x3)) = 1 + 2·x1 + x2 + x3   
POL(mark(x1)) = 2·x1   
POL(nil) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)


(2) Obligation:

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

a__from(X) → cons(X, from(s(X)))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__add(X1, X2) → add(X1, X2)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a__add(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(a__and(x1, x2)) = 2 + x1 + 2·x2   
POL(a__first(x1, x2)) = 2·x1 + x2   
POL(a__from(x1)) = 2·x1   
POL(a__if(x1, x2, x3)) = 2·x1 + x2 + x3   
POL(add(x1, x2)) = 1 + x1 + x2   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(first(x1, x2)) = 2·x1 + 2·x2   
POL(from(x1)) = x1   
POL(if(x1, x2, x3)) = 2·x1 + x2 + x3   
POL(mark(x1)) = 2·x1   
POL(nil) = 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__and(X1, X2) → and(X1, X2)
a__add(X1, X2) → add(X1, X2)


(4) Obligation:

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

a__from(X) → cons(X, from(s(X)))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a__first(x1, x2)) = 1 + 2·x1 + x2   
POL(a__from(x1)) = 2 + 2·x1   
POL(a__if(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(first(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(from(x1)) = 1 + x1   
POL(if(x1, x2, x3)) = 1 + 2·x1 + x2 + x3   
POL(mark(x1)) = 2·x1   
POL(nil) = 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:

mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(X1, X2)


(6) Obligation:

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

a__from(X) → cons(X, from(s(X)))
mark(from(X)) → a__from(X)
mark(nil) → nil

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a__from(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(from(x1)) = x1   
POL(mark(x1)) = 2·x1   
POL(nil) = 1   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(nil) → nil


(8) Obligation:

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

a__from(X) → cons(X, from(s(X)))
mark(from(X)) → a__from(X)

Q is empty.

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
a__from(x1)  =  a__from(x1)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  x1
s(x1)  =  x1
mark(x1)  =  mark(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > afrom1 > cons2

Status:
afrom1: multiset
cons2: [2,1]
mark1: multiset

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

a__from(X) → cons(X, from(s(X)))
mark(from(X)) → a__from(X)


(10) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(11) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(12) YES