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

(0) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))

The signature Sigma is {cond1, cond2, cond3}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

(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:

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND1(true, x, y) → GR(x, y)
COND2(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND2(true, x, y) → GR(add(x, y), 0)
COND2(true, x, y) → ADD(x, y)
COND2(true, x, y) → P(x)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND2(false, x, y) → EQ(x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(true, x, y) → GR(add(x, y), 0)
COND3(true, x, y) → ADD(x, y)
COND3(true, x, y) → P(x)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND3(false, x, y) → GR(add(x, y), 0)
COND3(false, x, y) → ADD(x, y)
COND3(false, x, y) → P(y)
GR(s(x), s(y)) → GR(x, y)
ADD(s(x), y) → ADD(x, y)
EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 11 less nodes.

(6) Complex Obligation (AND)

(7) 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:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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:

EQ(s(x), s(y)) → EQ(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

(11) 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.

(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:

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

(13) TRUE

(14) Obligation:

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

ADD(s(x), y) → ADD(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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:

ADD(s(x), y) → ADD(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

(18) Obligation:

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

ADD(s(x), y) → ADD(x, y)

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

(19) 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:

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

(20) TRUE

(21) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(22) 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.

(23) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(24) 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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

(25) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

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

(26) 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:

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

(27) TRUE

(28) Obligation:

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

COND2(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(29) 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.

(30) Obligation:

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

COND2(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(31) 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].

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)

(32) Obligation:

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

COND2(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(33) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, x, y) → COND1(gr(add(x, y), 0), p(x), y) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, 0, x0) → COND1(gr(x0, 0), p(0), x0)
COND2(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1)

(34) Obligation:

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND2(true, 0, x0) → COND1(gr(x0, 0), p(0), x0)
COND2(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(35) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, 0, x0) → COND1(gr(x0, 0), p(0), x0) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, 0, x0) → COND1(gr(x0, 0), 0, x0)

(36) Obligation:

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND2(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1)
COND2(true, 0, x0) → COND1(gr(x0, 0), 0, x0)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(37) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), x1) → COND1(true, p(s(x0)), x1)

(38) Obligation:

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND2(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND2(true, s(x0), x1) → COND1(true, p(s(x0)), x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(39) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), x1) → COND1(true, p(s(x0)), x1) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), x1) → COND1(true, x0, x1)

(40) Obligation:

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND2(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(41) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND1(true, x, y) → COND2(gr(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

COND1(true, 0, x0) → COND2(false, 0, x0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))

(42) Obligation:

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

COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND2(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, 0, x0) → COND2(false, 0, x0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(43) DependencyGraphProof (EQUIVALENT transformation)

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

(44) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y)
COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(45) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(true, x, y) → COND1(gr(add(x, y), 0), p(x), y) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, x0) → COND1(gr(x0, 0), p(0), x0)
COND3(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1)

(46) Obligation:

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

COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND3(true, 0, x0) → COND1(gr(x0, 0), p(0), x0)
COND3(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(47) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, x0) → COND1(gr(x0, 0), p(0), x0) at position [1] we obtained the following new rules [LPAR04]:

COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0)

(48) Obligation:

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

COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1)
COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(49) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), p(s(x0)), x1) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, s(x0), x1) → COND1(true, p(s(x0)), x1)

(50) Obligation:

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

COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND3(true, s(x0), x1) → COND1(true, p(s(x0)), x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(51) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, s(x0), x1) → COND1(true, p(s(x0)), x1) at position [1] we obtained the following new rules [LPAR04]:

COND3(true, s(x0), x1) → COND1(true, x0, x1)

(52) Obligation:

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

COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, x, y) → COND3(eq(x, y), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND3(true, s(x0), x1) → COND1(true, x0, x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(53) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(false, x, y) → COND3(eq(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, 0, 0) → COND3(true, 0, 0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND2(false, s(x0), 0) → COND3(false, s(x0), 0)
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))

(54) Obligation:

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

COND1(true, 0, x0) → COND2(false, 0, x0)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND2(false, 0, 0) → COND3(true, 0, 0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND2(false, s(x0), 0) → COND3(false, s(x0), 0)
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(55) DependencyGraphProof (EQUIVALENT transformation)

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

(56) Obligation:

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

COND2(false, 0, 0) → COND3(true, 0, 0)
COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0)
COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(57) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(true, 0, x0) → COND1(gr(x0, 0), 0, x0) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, 0) → COND1(false, 0, 0)
COND3(true, 0, s(x0)) → COND1(true, 0, s(x0))

(58) Obligation:

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

COND2(false, 0, 0) → COND3(true, 0, 0)
COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(true, 0, 0) → COND1(false, 0, 0)
COND3(true, 0, s(x0)) → COND1(true, 0, s(x0))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(59) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(60) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(61) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, x, y) → COND1(gr(add(x, y), 0), x, p(y)) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, x0) → COND1(gr(x0, 0), 0, p(x0))
COND3(false, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), s(x0), p(x1))

(62) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, 0, x0) → COND1(gr(x0, 0), 0, p(x0))
COND3(false, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), s(x0), p(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(63) DependencyGraphProof (EQUIVALENT transformation)

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

(64) Complex Obligation (AND)

(65) Obligation:

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

COND3(false, 0, x0) → COND1(gr(x0, 0), 0, p(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(66) 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.

(67) Obligation:

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

COND3(false, 0, x0) → COND1(gr(x0, 0), 0, p(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(68) 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].

add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

(69) Obligation:

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

COND3(false, 0, x0) → COND1(gr(x0, 0), 0, p(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

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

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

(70) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, 0, x0) → COND1(gr(x0, 0), 0, p(x0)) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, 0) → COND1(false, 0, p(0))
COND3(false, 0, s(x0)) → COND1(true, 0, p(s(x0)))

(71) Obligation:

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

COND1(true, 0, x0) → COND2(false, 0, x0)
COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, 0, 0) → COND1(false, 0, p(0))
COND3(false, 0, s(x0)) → COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

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

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

(72) DependencyGraphProof (EQUIVALENT transformation)

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

(73) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, 0, s(x0)) → COND1(true, 0, p(s(x0)))
COND1(true, 0, x0) → COND2(false, 0, x0)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

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

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

(74) 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.

(75) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, 0, s(x0)) → COND1(true, 0, p(s(x0)))
COND1(true, 0, x0) → COND2(false, 0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

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

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

(76) 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].

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

(77) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, 0, s(x0)) → COND1(true, 0, p(s(x0)))
COND1(true, 0, x0) → COND2(false, 0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(78) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(false, 0, s(x0)) → COND1(true, 0, p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND3(false, 0, s(x0)) → COND1(true, 0, x0)

(79) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND3(false, 0, s(x0)) → COND1(true, 0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(80) 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.

(81) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND3(false, 0, s(x0)) → COND1(true, 0, x0)

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

p(0)
p(s(x0))

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

(82) 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].

p(0)
p(s(x0))

(83) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND1(true, 0, x0) → COND2(false, 0, x0)
COND3(false, 0, s(x0)) → COND1(true, 0, x0)

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

(84) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND1(true, 0, x0) → COND2(false, 0, x0) we obtained the following new rules [LPAR04]:

COND1(true, 0, s(y_0)) → COND2(false, 0, s(y_0))

(85) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND3(false, 0, s(x0)) → COND1(true, 0, x0)
COND1(true, 0, s(y_0)) → COND2(false, 0, s(y_0))

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

(86) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND3(false, 0, s(x0)) → COND1(true, 0, x0) we obtained the following new rules [LPAR04]:

COND3(false, 0, s(s(y_0))) → COND1(true, 0, s(y_0))

(87) Obligation:

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

COND2(false, 0, s(x0)) → COND3(false, 0, s(x0))
COND1(true, 0, s(y_0)) → COND2(false, 0, s(y_0))
COND3(false, 0, s(s(y_0))) → COND1(true, 0, s(y_0))

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

(88) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND2(false, 0, s(x0)) → COND3(false, 0, s(x0)) we obtained the following new rules [LPAR04]:

COND2(false, 0, s(s(y_0))) → COND3(false, 0, s(s(y_0)))

(89) Obligation:

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

COND1(true, 0, s(y_0)) → COND2(false, 0, s(y_0))
COND3(false, 0, s(s(y_0))) → COND1(true, 0, s(y_0))
COND2(false, 0, s(s(y_0))) → COND3(false, 0, s(s(y_0)))

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

(90) 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:

• COND2(false, 0, s(s(y_0))) → COND3(false, 0, s(s(y_0)))
  The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3

• COND3(false, 0, s(s(y_0))) → COND1(true, 0, s(y_0))
  The graph contains the following edges 2 >= 2, 3 > 3

• COND1(true, 0, s(y_0)) → COND2(false, 0, s(y_0))
  The graph contains the following edges 2 >= 2, 3 >= 3

(91) TRUE

(92) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), s(x0), p(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(93) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(false, s(x0), x1) → COND1(gr(s(add(x0, x1)), 0), s(x0), p(x1)) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, s(x0), x1) → COND1(true, s(x0), p(x1))

(94) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(x0), x1) → COND1(true, s(x0), p(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(95) 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.

(96) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(x0), x1) → COND1(true, s(x0), p(x1))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(97) 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].

add(0, x0)
add(s(x0), x1)

(98) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(x0), x1) → COND1(true, s(x0), p(x1))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(99) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, s(x0), x1) → COND1(true, s(x0), p(x1)) at position [2] we obtained the following new rules [LPAR04]:

COND3(false, s(y0), 0) → COND1(true, s(y0), 0)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)

(100) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(y0), 0) → COND1(true, s(y0), 0)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(101) DependencyGraphProof (EQUIVALENT transformation)

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

(102) Obligation:

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

COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(103) 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.

(104) Obligation:

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

COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
p(0)
p(s(x0))

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

(105) 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].

p(0)
p(s(x0))

(106) Obligation:

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

COND2(true, s(x0), x1) → COND1(true, x0, x1)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(107) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(true, s(x0), x1) → COND1(true, x0, x1) we obtained the following new rules [LPAR04]:

COND2(true, s(z0), 0) → COND1(true, z0, 0)
COND2(true, s(z0), s(z1)) → COND1(true, z0, s(z1))

(108) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)
COND2(true, s(z0), 0) → COND1(true, z0, 0)
COND2(true, s(z0), s(z1)) → COND1(true, z0, s(z1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(109) DependencyGraphProof (EQUIVALENT transformation)

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

(110) Complex Obligation (AND)

(111) Obligation:

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

COND2(true, s(z0), 0) → COND1(true, z0, 0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(112) 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.

(113) Obligation:

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

COND2(true, s(z0), 0) → COND1(true, z0, 0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)

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

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(114) 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].

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

(115) Obligation:

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

COND2(true, s(z0), 0) → COND1(true, z0, 0)
COND1(true, s(x0), 0) → COND2(true, s(x0), 0)

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

(116) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND2(true, s(z0), 0) → COND1(true, z0, 0) we obtained the following new rules [LPAR04]:

COND2(true, s(s(y_0)), 0) → COND1(true, s(y_0), 0)

(117) Obligation:

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

COND1(true, s(x0), 0) → COND2(true, s(x0), 0)
COND2(true, s(s(y_0)), 0) → COND1(true, s(y_0), 0)

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

(118) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND1(true, s(x0), 0) → COND2(true, s(x0), 0) we obtained the following new rules [LPAR04]:

COND1(true, s(s(y_0)), 0) → COND2(true, s(s(y_0)), 0)

(119) Obligation:

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

COND2(true, s(s(y_0)), 0) → COND1(true, s(y_0), 0)
COND1(true, s(s(y_0)), 0) → COND2(true, s(s(y_0)), 0)

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

(120) 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:

• COND1(true, s(s(y_0)), 0) → COND2(true, s(s(y_0)), 0)
  The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3

• COND2(true, s(s(y_0)), 0) → COND1(true, s(y_0), 0)
  The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

(121) TRUE

(122) Obligation:

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

COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(x0), x1) → COND1(true, x0, x1)
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)
COND2(true, s(z0), s(z1)) → COND1(true, z0, s(z1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(123) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND3(true, s(x0), x1) → COND1(true, x0, x1) we obtained the following new rules [LPAR04]:

COND3(true, s(z0), s(z1)) → COND1(true, z0, s(z1))

(124) Obligation:

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

COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0)
COND2(true, s(z0), s(z1)) → COND1(true, z0, s(z1))
COND3(true, s(z0), s(z1)) → COND1(true, z0, s(z1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(125) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND3(false, s(y0), s(x0)) → COND1(true, s(y0), x0) we obtained the following new rules [LPAR04]:

COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))

(126) Obligation:

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

COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND2(true, s(z0), s(z1)) → COND1(true, z0, s(z1))
COND3(true, s(z0), s(z1)) → COND1(true, z0, s(z1))
COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(127) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND2(true, s(z0), s(z1)) → COND1(true, z0, s(z1)) we obtained the following new rules [LPAR04]:

COND2(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))

(128) Obligation:

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

COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(true, s(z0), s(z1)) → COND1(true, z0, s(z1))
COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))
COND2(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(129) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND3(true, s(z0), s(z1)) → COND1(true, z0, s(z1)) we obtained the following new rules [LPAR04]:

COND3(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))

(130) Obligation:

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

COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1))
COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))
COND2(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
COND3(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(131) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND2(false, s(x0), s(x1)) → COND3(eq(x0, x1), s(x0), s(x1)) we obtained the following new rules [LPAR04]:

COND2(false, s(x0), s(s(y_2))) → COND3(eq(x0, s(y_2)), s(x0), s(s(y_2)))
COND2(false, s(s(y_1)), s(x1)) → COND3(eq(s(y_1), x1), s(s(y_1)), s(x1))

(132) Obligation:

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

COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1))
COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))
COND2(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
COND3(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
COND2(false, s(x0), s(s(y_2))) → COND3(eq(x0, s(y_2)), s(x0), s(s(y_2)))
COND2(false, s(s(y_1)), s(x1)) → COND3(eq(s(y_1), x1), s(s(y_1)), s(x1))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(133) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND1(true, s(x0), s(x1)) → COND2(gr(x0, x1), s(x0), s(x1)) we obtained the following new rules [LPAR04]:

COND1(true, s(s(y_1)), s(x1)) → COND2(gr(s(y_1), x1), s(s(y_1)), s(x1))
COND1(true, s(x0), s(s(y_2))) → COND2(gr(x0, s(y_2)), s(x0), s(s(y_2)))

(134) Obligation:

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

COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))
COND2(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
COND3(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
COND2(false, s(x0), s(s(y_2))) → COND3(eq(x0, s(y_2)), s(x0), s(s(y_2)))
COND2(false, s(s(y_1)), s(x1)) → COND3(eq(s(y_1), x1), s(s(y_1)), s(x1))
COND1(true, s(s(y_1)), s(x1)) → COND2(gr(s(y_1), x1), s(s(y_1)), s(x1))
COND1(true, s(x0), s(s(y_2))) → COND2(gr(x0, s(y_2)), s(x0), s(s(y_2)))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

(135) 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:

• COND2(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
  The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

• COND1(true, s(s(y_1)), s(x1)) → COND2(gr(s(y_1), x1), s(s(y_1)), s(x1))
  The graph contains the following edges 2 >= 2, 3 >= 3

• COND1(true, s(x0), s(s(y_2))) → COND2(gr(x0, s(y_2)), s(x0), s(s(y_2)))
  The graph contains the following edges 2 >= 2, 3 >= 3

• COND2(false, s(x0), s(s(y_2))) → COND3(eq(x0, s(y_2)), s(x0), s(s(y_2)))
  The graph contains the following edges 2 >= 2, 3 >= 3

• COND2(false, s(s(y_1)), s(x1)) → COND3(eq(s(y_1), x1), s(s(y_1)), s(x1))
  The graph contains the following edges 2 >= 2, 3 >= 3

• COND3(false, s(x0), s(s(y_1))) → COND1(true, s(x0), s(y_1))
  The graph contains the following edges 2 >= 2, 3 > 3

• COND3(true, s(s(y_0)), s(x1)) → COND1(true, s(y_0), s(x1))
  The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

(136) TRUE