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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Quasi decreasingness of the given CTRS could not be shown:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 57 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) TransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) TransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) MRRProof [EQUIVALENT, 11 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) NonTerminationLoopProof [COMPLETE, 0 ms]
        (26) NO


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(0)
Obligation:
Conditional term rewrite system:
The TRS R consists of the following rules:

   gcd(x, 0) -> x
   gcd(0, x) -> x
   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)
   iadd(add(x, y)) -> tp2(x, y)

The conditional TRS C consists of the following conditional rules:

   gcd(z, y) -> gcd(x, y) <= iadd(z) -> tp2(x, y)
   gcd(y, z) -> gcd(x, y) <= iadd(z) -> tp2(x, y)


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(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   gcd(z, y) -> U1(iadd(z), y)
   U1(tp2(x, y), y) -> gcd(x, y)
   gcd(y, z) -> U2(iadd(z), y)
   U2(tp2(x, y), y) -> gcd(x, y)
   gcd(x, 0) -> x
   gcd(0, x) -> x
   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)
   iadd(add(x, y)) -> tp2(x, y)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(U1(x_1, x_2)) = x_1 + x_2
   POL(U2(x_1, x_2)) = x_1 + 2*x_2
   POL(add(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
   POL(gcd(x_1, x_2)) = 1 + 2*x_1 + x_2
   POL(iadd(x_1)) = 1 + x_1
   POL(s(x_1)) = x_1
   POL(tp2(x_1, x_2)) = 1 + 2*x_1 + x_2
   POL(u1(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   gcd(x, 0) -> x
   gcd(0, x) -> x
   iadd(add(x, y)) -> tp2(x, y)




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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   gcd(z, y) -> U1(iadd(z), y)
   U1(tp2(x, y), y) -> gcd(x, y)
   gcd(y, z) -> U2(iadd(z), y)
   U2(tp2(x, y), y) -> gcd(x, y)
   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.

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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GCD(z, y) -> U1^1(iadd(z), y)
   GCD(z, y) -> IADD(z)
   U1^1(tp2(x, y), y) -> GCD(x, y)
   GCD(y, z) -> U2^1(iadd(z), y)
   GCD(y, z) -> IADD(z)
   U2^1(tp2(x, y), y) -> GCD(x, y)
   IADD(s(z)) -> U1(iadd(z))
   IADD(s(z)) -> IADD(z)

The TRS R consists of the following rules:

   gcd(z, y) -> U1(iadd(z), y)
   U1(tp2(x, y), y) -> gcd(x, y)
   gcd(y, z) -> U2(iadd(z), y)
   U2(tp2(x, y), y) -> gcd(x, y)
   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
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(8)
Complex Obligation (AND)

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

   IADD(s(z)) -> IADD(z)

The TRS R consists of the following rules:

   gcd(z, y) -> U1(iadd(z), y)
   U1(tp2(x, y), y) -> gcd(x, y)
   gcd(y, z) -> U2(iadd(z), y)
   U2(tp2(x, y), y) -> gcd(x, y)
   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(10) 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.
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(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IADD(s(z)) -> IADD(z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(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:
*IADD(s(z)) -> IADD(z)
The graph contains the following edges 1 > 1


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(13)
YES

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

   U1^1(tp2(x, y), y) -> GCD(x, y)
   GCD(z, y) -> U1^1(iadd(z), y)
   GCD(y, z) -> U2^1(iadd(z), y)
   U2^1(tp2(x, y), y) -> GCD(x, y)

The TRS R consists of the following rules:

   gcd(z, y) -> U1(iadd(z), y)
   U1(tp2(x, y), y) -> gcd(x, y)
   gcd(y, z) -> U2(iadd(z), y)
   U2(tp2(x, y), y) -> gcd(x, y)
   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(15) 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.
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(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U1^1(tp2(x, y), y) -> GCD(x, y)
   GCD(z, y) -> U1^1(iadd(z), y)
   GCD(y, z) -> U2^1(iadd(z), y)
   U2^1(tp2(x, y), y) -> GCD(x, y)

The TRS R consists of the following rules:

   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(17) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule GCD(z, y) -> U1^1(iadd(z), y) at position [0] we obtained the following new rules [LPAR04]:

   (GCD(x0, y1) -> U1^1(tp2(0, x0), y1),GCD(x0, y1) -> U1^1(tp2(0, x0), y1))
   (GCD(s(x0), y1) -> U1^1(u1(iadd(x0)), y1),GCD(s(x0), y1) -> U1^1(u1(iadd(x0)), y1))


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

   U1^1(tp2(x, y), y) -> GCD(x, y)
   GCD(y, z) -> U2^1(iadd(z), y)
   U2^1(tp2(x, y), y) -> GCD(x, y)
   GCD(x0, y1) -> U1^1(tp2(0, x0), y1)
   GCD(s(x0), y1) -> U1^1(u1(iadd(x0)), y1)

The TRS R consists of the following rules:

   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(19) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule GCD(y, z) -> U2^1(iadd(z), y) at position [0] we obtained the following new rules [LPAR04]:

   (GCD(y0, x0) -> U2^1(tp2(0, x0), y0),GCD(y0, x0) -> U2^1(tp2(0, x0), y0))
   (GCD(y0, s(x0)) -> U2^1(u1(iadd(x0)), y0),GCD(y0, s(x0)) -> U2^1(u1(iadd(x0)), y0))


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(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U1^1(tp2(x, y), y) -> GCD(x, y)
   U2^1(tp2(x, y), y) -> GCD(x, y)
   GCD(x0, y1) -> U1^1(tp2(0, x0), y1)
   GCD(s(x0), y1) -> U1^1(u1(iadd(x0)), y1)
   GCD(y0, x0) -> U2^1(tp2(0, x0), y0)
   GCD(y0, s(x0)) -> U2^1(u1(iadd(x0)), y0)

The TRS R consists of the following rules:

   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(21) 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:

   GCD(y0, s(x0)) -> U2^1(u1(iadd(x0)), y0)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(GCD(x_1, x_2)) = x_1 + 2*x_2
   POL(U1^1(x_1, x_2)) = x_1 + x_2
   POL(U2^1(x_1, x_2)) = x_1 + x_2
   POL(iadd(x_1)) = x_1
   POL(s(x_1)) = 1 + 2*x_1
   POL(tp2(x_1, x_2)) = x_1 + x_2
   POL(u1(x_1)) = 1 + 2*x_1


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(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U1^1(tp2(x, y), y) -> GCD(x, y)
   U2^1(tp2(x, y), y) -> GCD(x, y)
   GCD(x0, y1) -> U1^1(tp2(0, x0), y1)
   GCD(s(x0), y1) -> U1^1(u1(iadd(x0)), y1)
   GCD(y0, x0) -> U2^1(tp2(0, x0), y0)

The TRS R consists of the following rules:

   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(23) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule U2^1(tp2(x, y), y) -> GCD(x, y) we obtained the following new rules [LPAR04]:

   (U2^1(tp2(0, y_0), y_0) -> GCD(0, y_0),U2^1(tp2(0, y_0), y_0) -> GCD(0, y_0))


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(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U1^1(tp2(x, y), y) -> GCD(x, y)
   GCD(x0, y1) -> U1^1(tp2(0, x0), y1)
   GCD(s(x0), y1) -> U1^1(u1(iadd(x0)), y1)
   GCD(y0, x0) -> U2^1(tp2(0, x0), y0)
   U2^1(tp2(0, y_0), y_0) -> GCD(0, y_0)

The TRS R consists of the following rules:

   iadd(y) -> tp2(0, y)
   iadd(s(z)) -> u1(iadd(z))
   u1(tp2(x, y)) -> tp2(s(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(25) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the right:

s = U1^1(tp2(x0, y1), y1) evaluates to  t =U1^1(tp2(0, x0), y1)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [x0 / 0, y1 / 0]

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Rewriting sequence

U1^1(tp2(0, 0), 0) -> GCD(0, 0)
 with rule U1^1(tp2(x, y), y) -> GCD(x, y) and matcher [x / 0, y / 0].

GCD(0, 0) -> U1^1(tp2(0, 0), 0)
with rule GCD(x0, y1) -> U1^1(tp2(0, x0), y1) at position [] and matcher [x0 / 0, y1 / 0]

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(26)
NO