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


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NO
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 disproven:

(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) MNOCProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) NonLoopProof [COMPLETE, 122 ms]
        (25) NO


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

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

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

The TRS R 2 is 
   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))

The signature Sigma is {f_2}
----------------------------------------

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

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

The set Q consists of the following terms:

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))


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

(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:

   F(tt, x) -> F(eq(toOne(x), s(0)), s(x))
   F(tt, x) -> EQ(toOne(x), s(0))
   F(tt, x) -> TOONE(x)
   EQ(s(x), s(y)) -> EQ(x, y)
   TOONE(s(s(x))) -> TOONE(s(x))

The TRS R consists of the following rules:

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

The set Q consists of the following terms:

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))

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 2 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   TOONE(s(s(x))) -> TOONE(s(x))

The TRS R consists of the following rules:

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

The set Q consists of the following terms:

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))

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:

   TOONE(s(s(x))) -> TOONE(s(x))

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

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))

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].

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))


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

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

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


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

(13)
YES

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

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

   EQ(s(x), s(y)) -> EQ(x, y)

The TRS R consists of the following rules:

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

The set Q consists of the following terms:

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))

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:

   EQ(s(x), s(y)) -> EQ(x, y)

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

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))

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].

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))


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

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

   EQ(s(x), s(y)) -> EQ(x, y)

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:
*EQ(s(x), s(y)) -> EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(20)
YES

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

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

   F(tt, x) -> F(eq(toOne(x), s(0)), s(x))

The TRS R consists of the following rules:

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

The set Q consists of the following terms:

   f(tt, x0)
   eq(s(x0), s(x1))
   eq(0, 0)
   toOne(s(s(x0)))
   toOne(s(0))

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

(22) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
----------------------------------------

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

   F(tt, x) -> F(eq(toOne(x), s(0)), s(x))

The TRS R consists of the following rules:

   f(tt, x) -> f(eq(toOne(x), s(0)), s(x))
   eq(s(x), s(y)) -> eq(x, y)
   eq(0, 0) -> tt
   toOne(s(s(x))) -> toOne(s(x))
   toOne(s(0)) -> s(0)

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

(24) NonLoopProof (COMPLETE)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 1, 
σ' = [ ], and μ' = [ ] on the rule
F(tt, s(s(zr0)))[zr0 / s(zr0)]^n[zr0 / 0] -> F(tt, s(s(s(zr0))))[zr0 / s(zr0)]^n[zr0 / 0]
This rule is correct for the QDP as the following derivation shows:

F(tt, s(s(zr0)))[zr0 / s(zr0)]^n[zr0 / 0] -> F(tt, s(s(s(zr0))))[zr0 / s(zr0)]^n[zr0 / 0]
    by Equivalence by Domain Renaming of the lhs with [zl0 / zr0]
    intermediate steps: Equiv DR (rhs) - Equiv DR (lhs)
    F(tt, s(s(zl1)))[zl1 / s(zl1)]^n[zl1 / 0] -> F(tt, s(s(s(zr1))))[zr1 / s(zr1)]^n[zr1 / 0]
        by Rewrite t with the rewrite sequence <Pos,Rule>: [([0],eq(s(x), s(y)) -> eq(x, y)), ([0],eq(0, 0) -> tt)]
        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs)
        F(tt, s(s(zl1)))[zr1 / s(zr1), zl1 / s(zl1)]^n[zr1 / 0, zl1 / 0] -> F(eq(s(0), s(0)), s(s(s(zr1))))[zr1 / s(zr1), zl1 / s(zl1)]^n[zr1 / 0, zl1 / 0]
            by Narrowing at position: [0,0]
            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
            F(tt, s(s(zs1)))[zs1 / s(zs1)]^n[zs1 / y0] -> F(eq(toOne(s(y0)), s(0)), s(s(s(zs1))))[zs1 / s(zs1)]^n[zs1 / y0]
                by Narrowing at position: [0,0]
                intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
                F(tt, x)[ ]^n[ ] -> F(eq(toOne(x), s(0)), s(x))[ ]^n[ ]
                    by Rule from TRS P

                intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
                toOne(s(s(x)))[x / s(x)]^n[ ] -> toOne(s(x))[ ]^n[ ]
                    by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x)]
                    toOne(s(s(x)))[ ]^n[ ] -> toOne(s(x))[ ]^n[ ]
                        by Rule from TRS R

            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
            toOne(s(0))[ ]^n[ ] -> s(0)[ ]^n[ ]
                by Rule from TRS R
----------------------------------------

(25)
NO