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:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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:

P(s(s(x))) → P(s(x))
INC(s(x)) → INC(x)
SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
SUMLIST(xs, y) → ISEMPTY(xs)
SUMLIST(xs, y) → ISZERO(head(xs))
SUMLIST(xs, y) → HEAD(xs)
SUMLIST(xs, y) → TAIL(xs)
SUMLIST(xs, y) → P(head(xs))
SUMLIST(xs, y) → INC(y)
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUM(xs) → SUMLIST(xs, 0)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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:

INC(s(x)) → INC(x)

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

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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].

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

(11) Obligation:

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

INC(s(x)) → INC(x)

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:

• INC(s(x)) → INC(x)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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:

P(s(s(x))) → P(s(x))

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

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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].

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

(18) Obligation:

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

P(s(s(x))) → P(s(x))

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:

• P(s(s(x))) → P(s(x))
  The graph contains the following edges 1 > 1

(20) TRUE

(21) Obligation:

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

SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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:

SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(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].

sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

(25) Obligation:

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

SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, true, y, xs, ys, x) → SUMLIST(xs, y) we obtained the following new rules [LPAR04]:

IF(false, true, z1, y_3, cons(y_5, y_6), y_7) → SUMLIST(y_3, z1)

(27) Obligation:

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

SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
IF(false, true, z1, y_3, cons(y_5, y_6), y_7) → SUMLIST(y_3, z1)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(28) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) at position [0] we obtained the following new rules [LPAR04]:

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(head(cons(x0, x1))), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))
SUMLIST(nil, y1) → IF(true, isZero(head(nil)), y1, tail(nil), cons(p(head(nil)), tail(nil)), inc(y1))

(29) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(head(cons(x0, x1))), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))
SUMLIST(nil, y1) → IF(true, isZero(head(nil)), y1, tail(nil), cons(p(head(nil)), tail(nil)), inc(y1))

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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:

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(head(cons(x0, x1))), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(32) 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.

(33) Obligation:

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

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(head(cons(x0, x1))), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(34) 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].

isEmpty(cons(x0, x1))
isEmpty(nil)

(35) Obligation:

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

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(head(cons(x0, x1))), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(36) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule SUMLIST(cons(x0, x1), y1) → IF(false, isZero(head(cons(x0, x1))), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1)) at position [1,0] we obtained the following new rules [LPAR04]:

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))

(37) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))

The TRS R consists of the following rules:

head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(38) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, tail(cons(x0, x1)), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1)) at position [3] we obtained the following new rules [LPAR04]:

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))

(39) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1))

The TRS R consists of the following rules:

head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(40) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), inc(y1)) at position [4,0,0] we obtained the following new rules [LPAR04]:

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), tail(cons(x0, x1))), inc(y1))

(41) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), tail(cons(x0, x1))), inc(y1))

The TRS R consists of the following rules:

head(cons(x, xs)) → x
isZero(0) → true
isZero(s(x)) → false
tail(cons(x, xs)) → xs
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(42) 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.

(43) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), tail(cons(x0, x1))), inc(y1))

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
tail(cons(x, xs)) → xs
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(44) 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].

head(cons(x0, x1))

(45) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), tail(cons(x0, x1))), inc(y1))

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
tail(cons(x, xs)) → xs
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(46) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), tail(cons(x0, x1))), inc(y1)) at position [4,1] we obtained the following new rules [LPAR04]:

SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), x1), inc(y1))

(47) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), x1), inc(y1))

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
tail(cons(x, xs)) → xs
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(48) 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.

(49) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), x1), inc(y1))

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(50) 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].

tail(cons(x0, x1))
tail(nil)

(51) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), x1), inc(y1))

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(52) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SUMLIST(cons(x0, x1), y1) → IF(false, isZero(x0), y1, x1, cons(p(x0), x1), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(p(0), y1), inc(y2))
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))

(53) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(p(0), y1), inc(y2))
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(54) 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.

(55) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(p(0), y1), inc(y2))
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(0) → 0

The set Q consists of the following terms:

isZero(0)
isZero(s(x0))
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)

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

(56) 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].

isZero(0)
isZero(s(x0))

(57) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(p(0), y1), inc(y2))
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(0) → 0

The set Q consists of the following terms:

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

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

(58) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(p(0), y1), inc(y2)) at position [4,0] we obtained the following new rules [LPAR04]:

SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(0, y1), inc(y2))

(59) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(0, y1), inc(y2))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(0) → 0

The set Q consists of the following terms:

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

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

(60) 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.

(61) Obligation:

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

IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(0, y1), inc(y2))

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

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

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

(62) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, true, y, xs, ys, x) → SUMLIST(xs, y) we obtained the following new rules [LPAR04]:

IF(false, true, z1, z0, cons(0, z0), y_0) → SUMLIST(z0, z1)

(63) Obligation:

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

IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(0, y1), inc(y2))
IF(false, true, z1, z0, cons(0, z0), y_0) → SUMLIST(z0, z1)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

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

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

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(false, true, z1, z0, cons(0, z0), y_0) → SUMLIST(z0, z1)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF(x₁, x₂, x₃, x₄, x₅, x₆)) = 1 + x₅   
POL(SUMLIST(x₁, x₂)) = 1 + x₁   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(false) = 0   
POL(inc(x₁)) = 0   
POL(p(x₁)) = 0   
POL(s(x₁)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented: none

(65) Obligation:

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

IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))
SUMLIST(cons(0, y1), y2) → IF(false, true, y2, y1, cons(0, y1), inc(y2))

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

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

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

(66) DependencyGraphProof (EQUIVALENT transformation)

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

(67) Obligation:

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

SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

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

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SUMLIST(cons(s(x0), y1), y2) → IF(false, false, y2, y1, cons(p(s(x0)), y1), inc(y2))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(SUMLIST(x1, x2)) =

/ 0 \ \ 0 /

+

/ 1 0 \ \ 0 0 /

·

x1

+

/ 0 0 \ \ 0 0 /

·

x2

POL(cons(x1, x2)) =

/ 0 \ \ 0 /

+

/ 0 1 \ \ 0 0 /

·

x1

+

/ 0 0 \ \ 0 0 /

·

x2

POL(s(x1)) =

/ 0 \ \ 1 /

+

/ 0 1 \ \ 0 1 /

·

x1

POL(IF(x1, x2, x3, x4, x5, x6)) =

/ 0 \ \ 0 /

+

/ 0 0 \ \ 0 0 /

·

x1

+

/ 0 0 \ \ 0 0 /

·

x2

+

/ 0 0 \ \ 0 0 /

·

x3

+

/ 0 0 \ \ 0 0 /

·

x4

+

/ 1 0 \ \ 0 1 /

·

x5

+

/ 0 0 \ \ 0 0 /

·

x6

POL(false) =

/ 0 \ \ 0 /

POL(p(x1)) =

/ 0 \ \ 0 /

+

/ 0 1 \ \ 1 0 /

·

x1

POL(inc(x1)) =

/ 0 \ \ 0 /

+

/ 0 0 \ \ 0 0 /

·

x1

POL(0) =

/ 0 \ \ 0 /

The following usable rules [FROCOS05] were oriented:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

(69) Obligation:

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

IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

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

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

(70) DependencyGraphProof (EQUIVALENT transformation)

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

(71) TRUE