YES
proof of Transformed_CSR_04_Ex4_4_Luc96b_GM.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 29 ms]
(4) QDP
(5) PisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a__f(g(X), Y) -> a__f(mark(X), f(g(X), Y))
   mark(f(X1, X2)) -> a__f(mark(X1), X2)
   mark(g(X)) -> g(mark(X))
   a__f(X1, X2) -> f(X1, X2)

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__F(g(X), Y) -> A__F(mark(X), f(g(X), Y))
   A__F(g(X), Y) -> MARK(X)
   MARK(f(X1, X2)) -> A__F(mark(X1), X2)
   MARK(f(X1, X2)) -> MARK(X1)
   MARK(g(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__f(g(X), Y) -> a__f(mark(X), f(g(X), Y))
   mark(f(X1, X2)) -> a__f(mark(X1), X2)
   mark(g(X)) -> g(mark(X))
   a__f(X1, X2) -> f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   A__F(g(X), Y) -> A__F(mark(X), f(g(X), Y))
   A__F(g(X), Y) -> MARK(X)
   MARK(f(X1, X2)) -> A__F(mark(X1), X2)
   MARK(f(X1, X2)) -> MARK(X1)
   MARK(g(X)) -> MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__F_2(x_1, x_2) ) = x_1 + 1
POL( mark_1(x_1) ) = 2x_1
POL( f_2(x_1, x_2) ) = 2x_1 + 1
POL( a__f_2(x_1, x_2) ) = 2x_1 + 2
POL( g_1(x_1) ) = 2x_1 + 1
POL( MARK_1(x_1) ) = x_1 + 1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(f(X1, X2)) -> a__f(mark(X1), X2)
   mark(g(X)) -> g(mark(X))
   a__f(X1, X2) -> f(X1, X2)
   a__f(g(X), Y) -> a__f(mark(X), f(g(X), Y))


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(4)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   a__f(g(X), Y) -> a__f(mark(X), f(g(X), Y))
   mark(f(X1, X2)) -> a__f(mark(X1), X2)
   mark(g(X)) -> g(mark(X))
   a__f(X1, X2) -> f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(6)
YES