YES Termination w.r.t. Q proof of AProVE_06_identity.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 17 ms)
↳4 QTRS
↳5 AAECC Innermost (⇔, 0 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 0 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 AND
    ↳11 QDP
        ↳12 UsableRulesProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QReductionProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QReductionProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPOrderProof (⇔, 38 ms)
        ↳31 QDP
        ↳32 PisEmptyProof (⇔, 0 ms)
        ↳33 YES

(0) Obligation:

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

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(d) = 0   
POL(double(x1)) = x1   
POL(f(x1, x2)) = x1 + x2   
POL(g(x1, x2)) = x1 + x2   
POL(h) = 0   
POL(half(x1)) = x1   
POL(id(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

id(x) → f(x, s(0))


(2) Obligation:

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

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(d) = 0   
POL(double(x1)) = x1   
POL(f(x1, x2)) = 2 + 2·x1 + x2   
POL(g(x1, x2)) = 2·x1 + x2   
POL(h) = 0   
POL(half(x1)) = x1   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(s(0), y) → y


(4) Obligation:

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

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

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

g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))

The TRS R 2 is

f(s(x), y) → f(half(s(x)), double(y))

The signature Sigma is {f}

(6) Obligation:

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

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

(7) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(8) Obligation:

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

G(d, s(x)) → G(d, x)
G(h, s(s(x))) → G(h, x)
DOUBLE(x) → G(d, x)
HALF(x) → G(h, x)
F(s(x), y) → F(half(s(x)), double(y))
F(s(x), y) → HALF(s(x))
F(s(x), y) → DOUBLE(y)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

G(h, s(s(x))) → G(h, x)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(12) 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.

(13) Obligation:

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

G(h, s(s(x))) → G(h, x)

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

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(14) 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].

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

(15) Obligation:

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

G(h, s(s(x))) → G(h, x)

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

(16) 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:

  • G(h, s(s(x))) → G(h, x)
    The graph contains the following edges 1 >= 1, 2 > 2

(17) YES

(18) Obligation:

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

G(d, s(x)) → G(d, x)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(19) 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.

(20) Obligation:

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

G(d, s(x)) → G(d, x)

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

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(21) 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].

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

(22) Obligation:

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

G(d, s(x)) → G(d, x)

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

(23) 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:

  • G(d, s(x)) → G(d, x)
    The graph contains the following edges 1 >= 1, 2 > 2

(24) YES

(25) Obligation:

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

F(s(x), y) → F(half(s(x)), double(y))

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(26) 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.

(27) Obligation:

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

F(s(x), y) → F(half(s(x)), double(y))

The TRS R consists of the following rules:

half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)

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

(28) 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)

(29) Obligation:

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

F(s(x), y) → F(half(s(x)), double(y))

The TRS R consists of the following rules:

half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)

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

(30) 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(half(s(x)), double(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(F(x1, x2)) = x1   
POL(d) = 0   
POL(double(x1)) = [2] + [4]x1   
POL(g(x1, x2)) = [1/2]x2   
POL(h) = [1]   
POL(half(x1)) = [1/2]x1   
POL(s(x1)) = [1] + [2]x1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

half(x) → g(h, x)
g(x, 0) → 0
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))

(31) Obligation:

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

half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))

The set Q consists of the following terms:

g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) YES