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:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

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:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

(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:

DOUBLE(s(x)) → DOUBLE(x)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
DOUBLELIST(cons(x, xs)) → DOUBLE(x)
DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))
DOUBLELIST(cons(x, xs)) → DEL(first(cons(x, xs)), cons(x, xs))
DOUBLELIST(cons(x, xs)) → FIRST(cons(x, xs))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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 4 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:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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].

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

(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:

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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:

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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].

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

(18) Obligation:

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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.

(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:

• DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF(false, x, y, xs) → DEL(x, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

(20) TRUE

(21) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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:

DOUBLE(s(x)) → DOUBLE(x)

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

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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].

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

(25) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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:

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

(27) TRUE

(28) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))

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].

double(0)
double(s(x0))
doublelist(nil)
doublelist(cons(x0, x1))

(32) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

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

(33) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs))) at position [0] we obtained the following new rules [LPAR04]:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(first(cons(x, xs)), x), first(cons(x, xs)), x, xs))

(34) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(first(cons(x, xs)), x), first(cons(x, xs)), x, xs))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

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

(35) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(first(cons(x, xs)), x), first(cons(x, xs)), x, xs)) at position [0,0,0] we obtained the following new rules [LPAR04]:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), first(cons(x, xs)), x, xs))

(36) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), first(cons(x, xs)), x, xs))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

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

(37) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), first(cons(x, xs)), x, xs)) at position [0,1] we obtained the following new rules [LPAR04]:

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

(38) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

The TRS R consists of the following rules:

first(cons(x, xs)) → x
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

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

(39) 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.

(40) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
del(x, nil) → nil
eq(0, s(y)) → false
eq(s(x), 0) → false

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))

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

(41) 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].

first(nil)
first(cons(x0, x1))

(42) Obligation:

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

DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
del(x, nil) → nil
eq(0, s(y)) → false
eq(s(x), 0) → false

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
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.

(43) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
DOUBLELIST(cons(x, xs)) → DOUBLELIST(if(eq(x, x), x, x, xs))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DOUBLELIST(x₁)) = x₁   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(del(x₁, x₂)) = x₂   
POL(eq(x₁, x₂)) = 0   
POL(false) = 0   
POL(if(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(nil) = 0   
POL(s(x₁)) = 0   
POL(true) = 0   

At least one of these decreasing rules is always used after the deleted DP:
if(true, x11'', y9'', xs4'') → xs4''


The following formula is valid:
x:sort[a19],xs:sort[a4].if'(eq(x , x ), x , x , xs )=true


The transformed set:
if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

(44) Complex Obligation (AND)

(45) Obligation:

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

eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
del(x, nil) → nil
eq(0, s(y)) → false
eq(s(x), 0) → false

The set Q consists of the following terms:

del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

Q is empty.

(49) Overlay + Local Confluence (EQUIVALENT transformation)

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

(50) Obligation:

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

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

(51) DependencyPairsProof (EQUIVALENT transformation)

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

(52) Obligation:

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

IF'(false, x18, y15, xs8) → DEL'(x18, xs8)
DEL'(x25, cons(y21, xs12)) → IF'(eq(x25, y21), x25, y21, xs12)
DEL'(x25, cons(y21, xs12)) → EQ(x25, y21)
EQ(s(x4), s(y3)) → EQ(x4, y3)
IF(false, x18, y15, xs8) → DEL(x18, xs8)
DEL(x25, cons(y21, xs12)) → IF(eq(x25, y21), x25, y21, xs12)
DEL(x25, cons(y21, xs12)) → EQ(x25, y21)
EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → AND(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x0, x2)
EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x1, x3)
EQUAL_SORT[A19](s(x0), s(x1)) → EQUAL_SORT[A19](x0, x1)

The TRS R consists of the following rules:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(53) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 3 less nodes.

(54) Complex Obligation (AND)

(55) Obligation:

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

EQUAL_SORT[A19](s(x0), s(x1)) → EQUAL_SORT[A19](x0, x1)

The TRS R consists of the following rules:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(56) 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.

(57) Obligation:

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

EQUAL_SORT[A19](s(x0), s(x1)) → EQUAL_SORT[A19](x0, x1)

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

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(58) 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].

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

(59) Obligation:

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

EQUAL_SORT[A19](s(x0), s(x1)) → EQUAL_SORT[A19](x0, x1)

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

(60) 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:

• EQUAL_SORT[A19](s(x0), s(x1)) → EQUAL_SORT[A19](x0, x1)
  The graph contains the following edges 1 > 1, 2 > 2

(61) TRUE

(62) Obligation:

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

EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x1, x3)
EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x0, x2)

The TRS R consists of the following rules:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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:

EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x1, x3)
EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x0, x2)

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

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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].

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

(66) Obligation:

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

EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x1, x3)
EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x0, x2)

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

(67) 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:

• EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x1, x3)
  The graph contains the following edges 1 > 1, 2 > 2

• EQUAL_SORT[A4](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A4](x0, x2)
  The graph contains the following edges 1 > 1, 2 > 2

(68) TRUE

(69) Obligation:

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

EQ(s(x4), s(y3)) → EQ(x4, y3)

The TRS R consists of the following rules:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(70) 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.

(71) Obligation:

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

EQ(s(x4), s(y3)) → EQ(x4, y3)

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

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(72) 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].

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

(73) Obligation:

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

EQ(s(x4), s(y3)) → EQ(x4, y3)

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

(74) 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(x4), s(y3)) → EQ(x4, y3)
  The graph contains the following edges 1 > 1, 2 > 2

(75) TRUE

(76) Obligation:

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

DEL(x25, cons(y21, xs12)) → IF(eq(x25, y21), x25, y21, xs12)
IF(false, x18, y15, xs8) → DEL(x18, xs8)

The TRS R consists of the following rules:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(77) 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.

(78) Obligation:

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

DEL(x25, cons(y21, xs12)) → IF(eq(x25, y21), x25, y21, xs12)
IF(false, x18, y15, xs8) → DEL(x18, xs8)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
eq(0, s(y32)) → false
eq(s(x45), 0) → false

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(79) 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].

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

(80) Obligation:

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

DEL(x25, cons(y21, xs12)) → IF(eq(x25, y21), x25, y21, xs12)
IF(false, x18, y15, xs8) → DEL(x18, xs8)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
eq(0, s(y32)) → false
eq(s(x45), 0) → false

The set Q consists of the following terms:

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

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

(81) 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:

• IF(false, x18, y15, xs8) → DEL(x18, xs8)
  The graph contains the following edges 2 >= 1, 4 >= 2

• DEL(x25, cons(y21, xs12)) → IF(eq(x25, y21), x25, y21, xs12)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

(82) TRUE

(83) Obligation:

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

DEL'(x25, cons(y21, xs12)) → IF'(eq(x25, y21), x25, y21, xs12)
IF'(false, x18, y15, xs8) → DEL'(x18, xs8)

The TRS R consists of the following rules:

if'(true, x11, y9, xs4) → true
if'(false, x18, y15, xs8) → del'(x18, xs8)
del'(x25, cons(y21, xs12)) → if'(eq(x25, y21), x25, y21, xs12)
del'(x32, nil) → false
eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
if(true, x11, y9, xs4) → xs4
if(false, x18, y15, xs8) → cons(y15, del(x18, xs8))
del(x25, cons(y21, xs12)) → if(eq(x25, y21), x25, y21, xs12)
del(x32, nil) → nil
eq(0, s(y32)) → false
eq(s(x45), 0) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a4](nil, nil) → true
equal_sort[a4](nil, cons(x0, x1)) → false
equal_sort[a4](cons(x0, x1), nil) → false
equal_sort[a4](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a4](x0, x2), equal_sort[a4](x1, x3))
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(84) 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.

(85) Obligation:

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

DEL'(x25, cons(y21, xs12)) → IF'(eq(x25, y21), x25, y21, xs12)
IF'(false, x18, y15, xs8) → DEL'(x18, xs8)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
eq(0, s(y32)) → false
eq(s(x45), 0) → false

The set Q consists of the following terms:

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
eq(0, 0)
eq(s(x0), s(x1))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
eq(0, s(x0))
eq(s(x0), 0)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

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

(86) 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].

if'(true, x0, x1, x2)
if'(false, x0, x1, x2)
del'(x0, cons(x1, x2))
del'(x0, nil)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
del(x0, cons(x1, x2))
del(x0, nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a4](nil, nil)
equal_sort[a4](nil, cons(x0, x1))
equal_sort[a4](cons(x0, x1), nil)
equal_sort[a4](cons(x0, x1), cons(x2, x3))
equal_sort[a19](0, 0)
equal_sort[a19](0, s(x0))
equal_sort[a19](s(x0), 0)
equal_sort[a19](s(x0), s(x1))
equal_sort[a37](witness_sort[a37], witness_sort[a37])

(87) Obligation:

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

DEL'(x25, cons(y21, xs12)) → IF'(eq(x25, y21), x25, y21, xs12)
IF'(false, x18, y15, xs8) → DEL'(x18, xs8)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(x4), s(y3)) → eq(x4, y3)
eq(0, s(y32)) → false
eq(s(x45), 0) → false

The set Q consists of the following terms:

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

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

(88) 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:

• IF'(false, x18, y15, xs8) → DEL'(x18, xs8)
  The graph contains the following edges 2 >= 1, 4 >= 2

• DEL'(x25, cons(y21, xs12)) → IF'(eq(x25, y21), x25, y21, xs12)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

(89) TRUE