YES Termination w.r.t. Q proof of Secret_05_TRS_tpa2.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 7 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPOrderProof (⇔, 40 ms)
        ↳20 QDP
        ↳21 QDPOrderProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 PisEmptyProof (⇔, 0 ms)
        ↳24 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The TRS R 2 is

f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The signature Sigma is {f}

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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:

-1(s(x), s(y)) → -1(x, y)
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) → P(-(s(x), y))
F(s(x), y) → -1(s(x), y)
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → -1(y, s(x))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → -1(x, s(y))
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → -1(s(y), x)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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:

-1(s(x), s(y)) → -1(x, y)

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

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

(10) QReductionProof (EQUIVALENT transformation)

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

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

-1(s(x), s(y)) → -1(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

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:

  • -1(s(x), s(y)) → -1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

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:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

(17) QReductionProof (EQUIVALENT transformation)

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

f(s(x0), x1)
f(x0, s(x1))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( F(x1, x2) ) = x1 + 2

POL( p(x1) ) = max{0, x1 - 2}

POL( -(x1, x2) ) = x1 + 1

POL( s(x1) ) = 2x1 + 2

POL( 0 ) = 0


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

-(s(x), s(y)) → -(x, y)
p(s(x)) → x
-(x, 0) → x

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( F(x1, x2) ) = 2x2 + 2

POL( p(x1) ) = max{0, x1 - 1}

POL( -(x1, x2) ) = x1

POL( s(x1) ) = 2x1 + 1

POL( 0 ) = 0


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

-(s(x), s(y)) → -(x, y)
p(s(x)) → x
-(x, 0) → x

(22) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) YES