/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPOrderProof [EQUIVALENT, 72 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) TRUE


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

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

   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

The TRS R 2 is 
   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))

The signature Sigma is {cond1_2, cond2_2}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))


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

(3) DependencyPairsProof (EQUIVALENT)
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:

   COND1(true, x) -> COND2(even(x), x)
   COND1(true, x) -> EVEN(x)
   COND2(true, x) -> COND1(neq(x, 0), div2(x))
   COND2(true, x) -> NEQ(x, 0)
   COND2(true, x) -> DIV2(x)
   COND2(false, x) -> COND1(neq(x, 0), p(x))
   COND2(false, x) -> NEQ(x, 0)
   COND2(false, x) -> P(x)
   NEQ(s(x), s(y)) -> NEQ(x, y)
   EVEN(s(s(x))) -> EVEN(x)
   DIV2(s(s(x))) -> DIV2(x)

The TRS R consists of the following rules:

   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 6 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   DIV2(s(s(x))) -> DIV2(x)

The TRS R consists of the following rules:

   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

(8) 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.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV2(s(s(x))) -> DIV2(x)

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

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV2(s(s(x))) -> DIV2(x)

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

(12) 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:
*DIV2(s(s(x))) -> DIV2(x)
The graph contains the following edges 1 > 1


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

(13)
YES

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EVEN(s(s(x))) -> EVEN(x)

The TRS R consists of the following rules:

   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

(15) 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.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EVEN(s(s(x))) -> EVEN(x)

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

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EVEN(s(s(x))) -> EVEN(x)

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

(19) 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:
*EVEN(s(s(x))) -> EVEN(x)
The graph contains the following edges 1 > 1


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

(20)
YES

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

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

   COND2(true, x) -> COND1(neq(x, 0), div2(x))
   COND1(true, x) -> COND2(even(x), x)
   COND2(false, x) -> COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

   cond1(true, x) -> cond2(even(x), x)
   cond2(true, x) -> cond1(neq(x, 0), div2(x))
   cond2(false, x) -> cond1(neq(x, 0), p(x))
   neq(0, 0) -> false
   neq(0, s(x)) -> true
   neq(s(x), 0) -> true
   neq(s(x), s(y)) -> neq(x, y)
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(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:

   COND2(true, x) -> COND1(neq(x, 0), div2(x))
   COND1(true, x) -> COND2(even(x), x)
   COND2(false, x) -> COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

   neq(0, 0) -> false
   neq(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))

The set Q consists of the following terms:

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)
   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

(24) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cond1(true, x0)
   cond2(true, x0)
   cond2(false, x0)


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

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

   COND2(true, x) -> COND1(neq(x, 0), div2(x))
   COND1(true, x) -> COND2(even(x), x)
   COND2(false, x) -> COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

   neq(0, 0) -> false
   neq(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))

The set Q consists of the following terms:

   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   COND1(true, x) -> COND2(even(x), x)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(COND1(x_1, x_2)) = [1/4]x_1 + [2]x_2
   POL(COND2(x_1, x_2)) = [2]x_2
   POL(div2(x_1)) = [1/2]x_1
   POL(even(x_1)) = 0
   POL(false) = 0
   POL(neq(x_1, x_2)) = [1/2]x_1
   POL(p(x_1)) = [1/4]x_1
   POL(s(x_1)) = [4] + [4]x_1
   POL(true) = [1/4]
The value of delta used in the strict ordering is 1/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   neq(0, 0) -> false
   neq(s(x), 0) -> true
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))
   p(0) -> 0
   p(s(x)) -> x


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, x) -> COND1(neq(x, 0), div2(x))
   COND2(false, x) -> COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

   neq(0, 0) -> false
   neq(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   even(0) -> true
   even(s(0)) -> false
   even(s(s(x))) -> even(x)
   div2(0) -> 0
   div2(s(0)) -> 0
   div2(s(s(x))) -> s(div2(x))

The set Q consists of the following terms:

   neq(0, 0)
   neq(0, s(x0))
   neq(s(x0), 0)
   neq(s(x0), s(y))
   even(0)
   even(s(0))
   even(s(s(x0)))
   div2(0)
   div2(s(0))
   div2(s(s(x0)))
   p(0)
   p(s(x0))

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

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

(29)
TRUE