YES Termination w.r.t. Q proof of Secret_07_TRS_5.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 TransformationProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 DependencyGraphProof (⇔, 0 ms)
        ↳11 TRUE
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 TransformationProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 DependencyGraphProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPOrderProof (⇔, 101 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPOrderProof (⇔, 10 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 TransformationProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 TransformationProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 TransformationProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 DependencyGraphProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 TransformationProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 TransformationProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 TransformationProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 DependencyGraphProof (⇔, 0 ms)
        ↳48 TRUE

(0) Obligation:

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

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

G(x, x, x) → G(c, d, e)
G(x, y, x) → G(c, d, e)
S(f(x, y)) → F(y, f(s(s(x)), a))
S(f(x, y)) → F(s(s(x)), a)
S(f(x, y)) → S(s(x))
S(f(x, y)) → S(x)
H(h(x, a), y) → H(h(a, y), h(a, x))
H(h(x, a), y) → H(a, y)
H(h(x, a), y) → H(a, x)
F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))
F(x, f(y, f(x, y))) → F(x, f(y, b))
F(x, f(y, f(x, y))) → F(y, b)
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(h(a, y), g(x, b, a)) → F(x, s(y))
F(h(a, y), g(x, b, a)) → S(y)
F(h(a, y), g(x, b, a)) → S(b)
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
H(f(x, s(y)), b) → G(y, a, f(s(x), a))
H(f(x, s(y)), b) → F(s(x), a)
H(f(x, s(y)), b) → S(x)
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(x, g(x, a, f(s(x), y))) → G(a, b, y)

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(6) UsableRulesProof (EQUIVALENT transformation)

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.

(7) Obligation:

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

F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))

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

(8) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, f(y, f(x, y))) → F(a, f(x, f(y, b))) we obtained the following new rules [LPAR04]:

F(a, f(b, f(a, b))) → F(a, f(a, f(b, b))) → F(a, f(b, f(a, b))) → F(a, f(a, f(b, b)))

(9) Obligation:

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

F(a, f(b, f(a, b))) → F(a, f(a, f(b, b)))

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

(10) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(11) TRUE

(12) Obligation:

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

S(f(x, y)) → S(x)
S(f(x, y)) → S(s(x))

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(13) UsableRulesProof (EQUIVALENT transformation)

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.

(14) Obligation:

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

S(f(x, y)) → S(x)
S(f(x, y)) → S(s(x))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(15) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule S(f(x, y)) → S(s(x)) at position [0] we obtained the following new rules [LPAR04]:

S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a))) → S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))
S(f(x0, y1)) → S(b) → S(f(x0, y1)) → S(b)

(16) Obligation:

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

S(f(x, y)) → S(x)
S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))
S(f(x0, y1)) → S(b)

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(18) Obligation:

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

S(f(x, y)) → S(x)
S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

Q is empty.
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.


S(f(x, y)) → S(x)
S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(S(x1)) = 0 +
[0,1]
·x1

POL(f(x1, x2)) =
/3\
\3/
 +
/00\
\11/
·x1 +
/11\
\00/
·x2

POL(s(x1)) =
/0\
\0/
 +
/03\
\10/
·x1

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

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

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

(20) Obligation:

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

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) YES

(23) Obligation:

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

H(h(x, a), y) → H(h(a, y), h(a, x))

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(24) UsableRulesProof (EQUIVALENT transformation)

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.

(25) Obligation:

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

H(h(x, a), y) → H(h(a, y), h(a, x))

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

(26) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


H(h(x, a), y) → H(h(a, y), h(a, x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(H(x1, x2)) = [4]x1 + [2]x2   
POL(a) = [1/2]   
POL(h(x1, x2)) = [1/4]x1 + [1/2]x2   
The value of delta used in the strict ordering is 1/4.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(27) Obligation:

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES

(30) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(31) UsableRulesProof (EQUIVALENT transformation)

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.

(32) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(33) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y)) we obtained the following new rules [LPAR04]:

F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1)) → F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))

(34) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(35) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b)) we obtained the following new rules [LPAR04]:

F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b)) → F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))

(36) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(37) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, g(x, a, f(s(x), y))) → H(x, b) we obtained the following new rules [LPAR04]:

F(a, g(a, a, f(s(a), x1))) → H(a, b) → F(a, g(a, a, f(s(a), x1))) → H(a, b)

(38) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))
F(a, g(a, a, f(s(a), x1))) → H(a, b)

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(39) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(40) Obligation:

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

F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(41) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b)) at position [1] we obtained the following new rules [LPAR04]:

F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b) → F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)

(42) Obligation:

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

F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(43) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1)) we obtained the following new rules [LPAR04]:

F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a)) → F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a))

(44) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)
F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(45) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a))) we obtained the following new rules [LPAR04]:

H(f(a, s(a)), b) → F(a, g(a, a, f(s(a), a))) → H(f(a, s(a)), b) → F(a, g(a, a, f(s(a), a)))

(46) Obligation:

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

F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)
F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a))
H(f(a, s(a)), b) → F(a, g(a, a, f(s(a), a)))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
g(x, y, x) → g(c, d, e)
h(h(x, a), y) → h(h(a, y), h(a, x))

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

(47) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(48) TRUE