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

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, 0), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), x, p(y))
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, 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, 0), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), x, p(y))
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, 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))
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)
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, 0), x, y)
COND1(true, x, y) → GR(x, 0)
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(gr(y, 0), x, y)
COND2(false, x, y) → GR(y, 0)
COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND3(true, x, y) → GR(add(x, y), 0)
COND3(true, x, y) → ADD(x, y)
COND3(true, x, y) → P(y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND3(false, x, y) → GR(add(x, y), 0)
COND3(false, x, y) → ADD(x, y)
GR(s(x), s(y)) → GR(x, y)
ADD(s(x), y) → ADD(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, 0), x, y)
cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)
cond3(true, x, y) → cond1(gr(add(x, y), 0), x, p(y))
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, 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))
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)
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 3 SCCs with 10 less nodes.

(6) Complex Obligation (AND)

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

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)
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)
p(0)
p(s(x0))

(11) 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.

(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:

• ADD(s(x), y) → ADD(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:

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

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

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)
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)
p(0)
p(s(x0))

(18) 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.

(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:

• GR(s(x), s(y)) → GR(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:

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

The TRS R consists of the following rules:

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

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

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)
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)

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

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)
p(0)
p(s(x0))

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

(26) Narrowing (EQUIVALENT transformation)

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)

(27) 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)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)

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

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)
p(0)
p(s(x0))

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

(28) Narrowing (EQUIVALENT transformation)

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

COND2(true, 0, y1) → COND1(gr(add(0, y1), 0), 0, y1)
COND2(true, s(x0), y1) → COND1(gr(add(s(x0), y1), 0), x0, y1)

(29) Obligation:

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

COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, 0, y1) → COND1(gr(add(0, y1), 0), 0, y1)
COND2(true, s(x0), y1) → COND1(gr(add(s(x0), y1), 0), x0, y1)

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

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)
p(0)
p(s(x0))

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

(30) DependencyGraphProof (EQUIVALENT transformation)

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

(31) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(gr(add(s(x0), y1), 0), x0, y1)

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

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)
p(0)
p(s(x0))

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

(32) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), y1) → COND1(gr(add(s(x0), y1), 0), x0, y1) at position [0,0] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), y1) → COND1(gr(s(add(x0, y1)), 0), x0, y1)

(33) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(gr(s(add(x0, y1)), 0), x0, y1)

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

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)
p(0)
p(s(x0))

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

(34) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), y1) → COND1(gr(s(add(x0, y1)), 0), x0, y1) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), y1) → COND1(true, x0, y1)

(35) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, x, y) → COND3(gr(y, 0), x, y)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)

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

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)
p(0)
p(s(x0))

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

(36) Narrowing (EQUIVALENT transformation)

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

COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))

(37) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(false, x, y) → COND1(gr(add(x, y), 0), x, y)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, 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

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)
p(0)
p(s(x0))

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

(38) Instantiation (EQUIVALENT transformation)

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

COND3(false, z0, 0) → COND1(gr(add(z0, 0), 0), z0, 0)

(39) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(false, y0, 0) → COND3(false, y0, 0)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(false, z0, 0) → COND1(gr(add(z0, 0), 0), z0, 0)

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

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)
p(0)
p(s(x0))

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

(40) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, y0, 0) → COND3(false, y0, 0) we obtained the following new rules [LPAR04]:

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

(41) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(false, z0, 0) → COND1(gr(add(z0, 0), 0), z0, 0)
COND2(false, 0, 0) → COND3(false, 0, 0)

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

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)
p(0)
p(s(x0))

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

(42) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, z0, 0) → COND1(gr(add(z0, 0), 0), z0, 0) at position [0] we obtained the following new rules [LPAR04]:

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

(43) Obligation:

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

COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND2(false, 0, 0) → COND3(false, 0, 0)
COND3(false, 0, 0) → COND1(gr(0, 0), 0, 0)
COND3(false, s(x0), 0) → COND1(gr(s(add(x0, 0)), 0), s(x0), 0)

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

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)
p(0)
p(s(x0))

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

(44) DependencyGraphProof (EQUIVALENT transformation)

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

(45) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND2(false, y0, s(x0)) → COND3(true, y0, s(x0))
COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)

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

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)
p(0)
p(s(x0))

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

(46) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, y0, s(x0)) → COND3(true, y0, s(x0)) we obtained the following new rules [LPAR04]:

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

(47) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, x, y) → COND1(gr(add(x, y), 0), x, p(y))
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(false, 0, s(x1)) → COND3(true, 0, 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

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)
p(0)
p(s(x0))

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

(48) Instantiation (EQUIVALENT transformation)

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

COND3(true, 0, s(z0)) → COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))

(49) Obligation:

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

COND1(true, 0, y1) → COND2(false, 0, y1)
COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)
COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, 0, s(z0)) → COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))

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

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)
p(0)
p(s(x0))

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

(50) DependencyGraphProof (EQUIVALENT transformation)

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

(51) Complex Obligation (AND)

(52) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, 0, s(z0)) → COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))
COND1(true, 0, y1) → COND2(false, 0, y1)

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

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)
p(0)
p(s(x0))

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

(53) 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.

(54) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND3(true, 0, s(z0)) → COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))
COND1(true, 0, y1) → COND2(false, 0, y1)

The TRS R consists of the following rules:

add(0, x) → x
gr(0, x) → false
gr(s(x), 0) → true
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)
p(0)
p(s(x0))

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

(55) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, s(z0)) → COND1(gr(add(0, s(z0)), 0), 0, p(s(z0))) at position [0,0] we obtained the following new rules [LPAR04]:

COND3(true, 0, s(z0)) → COND1(gr(s(z0), 0), 0, p(s(z0)))

(56) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(gr(s(z0), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

add(0, x) → x
gr(0, x) → false
gr(s(x), 0) → true
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)
p(0)
p(s(x0))

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

(57) 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.

(58) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(gr(s(z0), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

gr(s(x), 0) → true
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)
p(0)
p(s(x0))

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

(59) 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)

(60) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(gr(s(z0), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

gr(s(x), 0) → true
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.

(61) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, s(z0)) → COND1(gr(s(z0), 0), 0, p(s(z0))) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, s(z0)) → COND1(true, 0, p(s(z0)))

(62) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(true, 0, p(s(z0)))

The TRS R consists of the following rules:

gr(s(x), 0) → true
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.

(63) 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.

(64) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(true, 0, p(s(z0)))

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.

(65) 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))

(66) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(true, 0, p(s(z0)))

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.

(67) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, s(z0)) → COND1(true, 0, p(s(z0))) at position [2] we obtained the following new rules [LPAR04]:

COND3(true, 0, s(z0)) → COND1(true, 0, z0)

(68) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(true, 0, z0)

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.

(69) 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.

(70) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(true, 0, z0)

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.

(71) 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))

(72) Obligation:

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

COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
COND1(true, 0, y1) → COND2(false, 0, y1)
COND3(true, 0, s(z0)) → COND1(true, 0, z0)

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

(73) 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:

• COND3(true, 0, s(z0)) → COND1(true, 0, z0)
  The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

• COND1(true, 0, y1) → COND2(false, 0, y1)
  The graph contains the following edges 2 >= 2, 3 >= 3

• COND2(false, 0, s(x1)) → COND3(true, 0, s(x1))
  The graph contains the following edges 2 >= 2, 3 >= 3

(74) TRUE

(75) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)

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

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)
p(0)
p(s(x0))

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

(76) 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.

(77) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)

R is empty.
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)
p(0)
p(s(x0))

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

(78) 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))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

(79) Obligation:

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

COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
COND2(true, s(x0), y1) → COND1(true, x0, y1)

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

(80) 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(x0), y1) → COND1(true, x0, y1)
  The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

• COND1(true, s(x0), y1) → COND2(true, s(x0), y1)
  The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3

(81) TRUE