YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 26 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) MRRProof [EQUIVALENT, 28 ms]
        (19) QDP
        (20) MNOCProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QDPOrderProof [EQUIVALENT, 18 ms]
        (25) QDP
        (26) PisEmptyProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) MRRProof [EQUIVALENT, 36 ms]
        (32) QDP
        (33) MNOCProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) UsableRulesProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) QDPOrderProof [EQUIVALENT, 106 ms]
        (38) QDP
        (39) DependencyGraphProof [EQUIVALENT, 0 ms]
        (40) TRUE


----------------------------------------

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

   p(0(x1)) -> 0(s(s(p(x1))))
   p(s(x1)) -> x1
   p(p(s(x1))) -> p(x1)
   f(s(x1)) -> g(s(x1))
   g(x1) -> i(s(half(x1)))
   i(x1) -> f(p(x1))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   0(x1) -> x1
   rd(0(x1)) -> 0(0(0(0(0(0(rd(x1)))))))

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(0(x1)) -> 0^1(s(s(p(x1))))
   P(0(x1)) -> P(x1)
   P(p(s(x1))) -> P(x1)
   F(s(x1)) -> G(s(x1))
   G(x1) -> I(s(half(x1)))
   G(x1) -> HALF(x1)
   I(x1) -> F(p(x1))
   I(x1) -> P(x1)
   HALF(0(x1)) -> 0^1(s(s(half(x1))))
   HALF(0(x1)) -> HALF(x1)
   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))
   HALF(s(s(x1))) -> P(p(s(s(x1))))
   HALF(s(s(x1))) -> P(s(s(x1)))
   RD(0(x1)) -> 0^1(0(0(0(0(0(rd(x1)))))))
   RD(0(x1)) -> 0^1(0(0(0(0(rd(x1))))))
   RD(0(x1)) -> 0^1(0(0(0(rd(x1)))))
   RD(0(x1)) -> 0^1(0(0(rd(x1))))
   RD(0(x1)) -> 0^1(0(rd(x1)))
   RD(0(x1)) -> 0^1(rd(x1))
   RD(0(x1)) -> RD(x1)

The TRS R consists of the following rules:

   p(0(x1)) -> 0(s(s(p(x1))))
   p(s(x1)) -> x1
   p(p(s(x1))) -> p(x1)
   f(s(x1)) -> g(s(x1))
   g(x1) -> i(s(half(x1)))
   i(x1) -> f(p(x1))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   0(x1) -> x1
   rd(0(x1)) -> 0(0(0(0(0(0(rd(x1)))))))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 12 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

----------------------------------------

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   RD(0(x1)) -> RD(x1)

The TRS R consists of the following rules:

   p(0(x1)) -> 0(s(s(p(x1))))
   p(s(x1)) -> x1
   p(p(s(x1))) -> p(x1)
   f(s(x1)) -> g(s(x1))
   g(x1) -> i(s(half(x1)))
   i(x1) -> f(p(x1))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   0(x1) -> x1
   rd(0(x1)) -> 0(0(0(0(0(0(rd(x1)))))))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   RD(0(x1)) -> RD(x1)

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

(8) QDPSizeChangeProof (EQUIVALENT)
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:
*RD(0(x1)) -> RD(x1)
The graph contains the following edges 1 > 1


----------------------------------------

(9)
YES

----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(p(s(x1))) -> P(x1)
   P(0(x1)) -> P(x1)

The TRS R consists of the following rules:

   p(0(x1)) -> 0(s(s(p(x1))))
   p(s(x1)) -> x1
   p(p(s(x1))) -> p(x1)
   f(s(x1)) -> g(s(x1))
   g(x1) -> i(s(half(x1)))
   i(x1) -> f(p(x1))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   0(x1) -> x1
   rd(0(x1)) -> 0(0(0(0(0(0(rd(x1)))))))

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

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(p(s(x1))) -> P(x1)
   P(0(x1)) -> P(x1)

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

(13) QDPSizeChangeProof (EQUIVALENT)
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:
*P(p(s(x1))) -> P(x1)
The graph contains the following edges 1 > 1


*P(0(x1)) -> P(x1)
The graph contains the following edges 1 > 1


----------------------------------------

(14)
YES

----------------------------------------

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))
   HALF(0(x1)) -> HALF(x1)

The TRS R consists of the following rules:

   p(0(x1)) -> 0(s(s(p(x1))))
   p(s(x1)) -> x1
   p(p(s(x1))) -> p(x1)
   f(s(x1)) -> g(s(x1))
   g(x1) -> i(s(half(x1)))
   i(x1) -> f(p(x1))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   0(x1) -> x1
   rd(0(x1)) -> 0(0(0(0(0(0(rd(x1)))))))

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

(16) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))
   HALF(0(x1)) -> HALF(x1)

The TRS R consists of the following rules:

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   p(p(s(x1))) -> p(x1)
   0(x1) -> x1

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

(18) MRRProof (EQUIVALENT)
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:

   HALF(0(x1)) -> HALF(x1)

Strictly oriented rules of the TRS R:

   0(x1) -> x1

Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = 2 + 2*x_1
   POL(HALF(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = x_1


----------------------------------------

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   p(p(s(x1))) -> p(x1)

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

(20) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   p(p(s(x1))) -> p(x1)

The set Q consists of the following terms:

   p(s(x0))
   p(0(x0))

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

(22) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))

The set Q consists of the following terms:

   p(s(x0))
   p(0(x0))

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

(24) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   HALF(s(s(x1))) -> HALF(p(p(s(s(x1)))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( HALF_1(x_1) ) = 2x_1 + 2
POL( p_1(x_1) ) = max{0, x_1 - 2}
POL( s_1(x_1) ) = 2x_1 + 2
POL( 0_1(x_1) ) = max{0, -2}

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

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))


----------------------------------------

(25)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))

The set Q consists of the following terms:

   p(s(x0))
   p(0(x0))

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

(26) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(27)
YES

----------------------------------------

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   I(x1) -> F(p(x1))
   F(s(x1)) -> G(s(x1))
   G(x1) -> I(s(half(x1)))

The TRS R consists of the following rules:

   p(0(x1)) -> 0(s(s(p(x1))))
   p(s(x1)) -> x1
   p(p(s(x1))) -> p(x1)
   f(s(x1)) -> g(s(x1))
   g(x1) -> i(s(half(x1)))
   i(x1) -> f(p(x1))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   0(x1) -> x1
   rd(0(x1)) -> 0(0(0(0(0(0(rd(x1)))))))

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

(29) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   I(x1) -> F(p(x1))
   F(s(x1)) -> G(s(x1))
   G(x1) -> I(s(half(x1)))

The TRS R consists of the following rules:

   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   p(p(s(x1))) -> p(x1)
   0(x1) -> x1

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

(31) MRRProof (EQUIVALENT)
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 rules of the TRS R:

   0(x1) -> x1

Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = 2 + x_1
   POL(F(x_1)) = x_1
   POL(G(x_1)) = x_1
   POL(I(x_1)) = x_1
   POL(half(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = x_1


----------------------------------------

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   I(x1) -> F(p(x1))
   F(s(x1)) -> G(s(x1))
   G(x1) -> I(s(half(x1)))

The TRS R consists of the following rules:

   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   p(p(s(x1))) -> p(x1)

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

(33) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   I(x1) -> F(p(x1))
   F(s(x1)) -> G(s(x1))
   G(x1) -> I(s(half(x1)))

The TRS R consists of the following rules:

   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   p(p(s(x1))) -> p(x1)

The set Q consists of the following terms:

   half(0(x0))
   half(s(s(x0)))
   p(s(x0))
   p(0(x0))

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

(35) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   I(x1) -> F(p(x1))
   F(s(x1)) -> G(s(x1))
   G(x1) -> I(s(half(x1)))

The TRS R consists of the following rules:

   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))

The set Q consists of the following terms:

   half(0(x0))
   half(s(s(x0)))
   p(s(x0))
   p(0(x0))

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

(37) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   F(s(x1)) -> G(s(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(I(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(p(x_1)) =  	[[0A], [-I], [0A]] 	 +  	[[-I, 0A, 0A], [-I, 0A, -I], [-I, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(s(x_1)) =  	[[1A], [0A], [0A]] 	 +  	[[1A, 1A, 1A], [-I, 0A, 0A], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(G(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(half(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, 0A, 0A], [-I, 0A, 0A], [-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(0(x_1)) =  	[[0A], [0A], [-I]] 	 +  	[[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] 	* 	x_1
>>>


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

   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))
   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))


----------------------------------------

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   I(x1) -> F(p(x1))
   G(x1) -> I(s(half(x1)))

The TRS R consists of the following rules:

   half(0(x1)) -> 0(s(s(half(x1))))
   half(s(s(x1))) -> s(half(p(p(s(s(x1))))))
   p(s(x1)) -> x1
   p(0(x1)) -> 0(s(s(p(x1))))

The set Q consists of the following terms:

   half(0(x0))
   half(s(s(x0)))
   p(s(x0))
   p(0(x0))

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

(39) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(40)
TRUE