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:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

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:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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:

MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
MAX(cons(x, cons(y, xs))) → GE(x, y)
IF1(true, x, y, xs) → MAX(cons(x, xs))
IF1(false, x, y, xs) → MAX(cons(y, xs))
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF2(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
SORT(cons(x, xs)) → MAX(cons(x, xs))
SORT(cons(x, xs)) → SORT(del(max(cons(x, xs)), cons(x, xs)))
SORT(cons(x, xs)) → DEL(max(cons(x, xs)), cons(x, xs))
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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 5 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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:

GE(s(x), s(y)) → GE(x, y)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(11) Obligation:

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

GE(s(x), s(y)) → GE(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:

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

(13) YES

(14) 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:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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:

EQ(s(x), s(y)) → EQ(x, y)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(18) 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.

(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:

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

(20) YES

(21) Obligation:

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

IF2(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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:

IF2(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF2(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:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(25) Obligation:

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

IF2(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF2(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.

(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:

• DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2(false, x, y, xs) → DEL(x, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

(27) YES

(28) Obligation:

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

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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:

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(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].

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))

(32) Obligation:

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

IF1(true, x, y, xs) → MAX(cons(x, xs))
MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)

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

POL(0) = 0   
POL(IF1(x₁, x₂, x₃, x₄)) = 1 + x₂ + x₃ + x₄   
POL(MAX(x₁)) = x₁   
POL(cons(x₁, x₂)) = 1 + x₁ + x₂   
POL(false) = 0   
POL(ge(x₁, x₂)) = 0   
POL(s(x₁)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

IF1(true, x, y, xs) → MAX(cons(x, xs))
IF1(false, x, y, xs) → MAX(cons(y, xs))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(35) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(36) TRUE

(37) Obligation:

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

SORT(cons(x, xs)) → SORT(del(max(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(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)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(38) 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.

(39) Obligation:

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

SORT(cons(x, xs)) → SORT(del(max(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, cons(y, xs)) → if2(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)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(nil)
sort(cons(x0, x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(40) 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].

sort(nil)
sort(cons(x0, x1))

(41) Obligation:

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

SORT(cons(x, xs)) → SORT(del(max(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, cons(y, xs)) → if2(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)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(42) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
SORT(cons(x, xs)) → SORT(del(max(cons(x, xs)), cons(x, xs)))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(SORT(x₁)) = x₁   
POL(cons(x₁, x₂)) = 2 + x₁ + x₂   
POL(del(x₁, x₂)) = x₂   
POL(eq(x₁, x₂)) = 1   
POL(false) = 1   
POL(ge(x₁, x₂)) = 1   
POL(if1(x₁, x₂, x₃, x₄)) = 3 + x₁ + 2·x₂ + 2·x₃ + 2·x₄   
POL(if2(x₁, x₂, x₃, x₄)) = 2 + x₃ + x₄   
POL(max(x₁)) = 2·x₁   
POL(nil) = 0   
POL(s(x₁)) = 0   
POL(true) = 1   

At least one of these decreasing rules is always used after the deleted DP:
if2(true, x1227, y857, xs567) → xs567


The following formula is valid:
z0:sort[a34].(¬(z0 =nil)→del'(max(z0 ), z0 )=true)


The transformed set:
del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true


The proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, x2, x3, x122, y85, xs56, x136, y95, y37, xs25, x150, x54, x12, y7, xs4, y56, x94, x108, y75, x177, x191, x205, y141, x26, y17, x40, y27, xs11, xs18]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or 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[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a34](cons(x0, x1), cons(x2, x3)) -> equal_sort[a34](x0, x2) and equal_sort[a34](x1, x3)
   equal_sort[a34](cons(x0, x1), nil) -> false
   equal_sort[a34](nil, cons(x0, x1)) -> false
   equal_sort[a34](nil, nil) -> true
   equal_sort[a60](witness_sort[a60], witness_sort[a60]) -> true
   if2'(true, x122, y85, xs56) -> true
   if2'(false, x136, y95, cons(y37, xs25)) -> if2'(eq(x136, y37), x136, y37, xs25)
   if2'(false, x136, y95, nil) -> false
   del'(x150, nil) -> false
   equal_bool(eq(x54, y37), true) -> true | del'(x54, cons(y37, xs25)) -> true
   equal_bool(eq(x54, y37), true) -> false | del'(x54, cons(y37, xs25)) -> del'(x54, xs25)
   max(cons(x, nil)) -> x
   max(nil) -> 0
   equal_bool(ge(x12, y7), true) -> true | max(cons(x12, cons(y7, xs4))) -> max(cons(x12, xs4))
   equal_bool(ge(x12, y7), true) -> false | max(cons(x12, cons(y7, xs4))) -> max(cons(y7, xs4))
   eq(0, 0) -> true
   eq(0, s(y56)) -> false
   eq(s(x94), 0) -> false
   eq(s(x108), s(y75)) -> eq(x108, y75)
   if2(true, x122, y85, xs56) -> xs56
   if2(false, x136, y95, cons(y37, xs25)) -> cons(y95, if2(eq(x136, y37), x136, y37, xs25))
   if2(false, x136, y95, nil) -> cons(y95, nil)
   del(x150, nil) -> nil
   equal_bool(eq(x54, y37), true) -> true | del(x54, cons(y37, xs25)) -> xs25
   equal_bool(eq(x54, y37), true) -> false | del(x54, cons(y37, xs25)) -> cons(y37, del(x54, xs25))
   ge(0, 0) -> true
   ge(s(x177), 0) -> true
   ge(0, s(x191)) -> false
   ge(s(x205), s(y141)) -> ge(x205, y141)
   if1(true, x26, y17, nil) -> x26
   if1(true, x26, y17, cons(y7, xs4)) -> if1(ge(x26, y7), x26, y7, xs4)
   if1(false, x40, y27, nil) -> y27
   if1(false, x40, y27, cons(y7, xs4)) -> if1(ge(y27, y7), y27, y7, xs4)
   if1(true, x26, y17, xs11) -> 0
   if1(false, x40, y27, xs18) -> 0

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, x2, x3, x122, y85, xs56, x136, y95, y37, xs25, x150, x54, x12, y7, xs4, y56, x94, x108, y75, x177, x191, x205, y141, x26, y17, x40, y27, xs11, xs18]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or 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[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a34](cons(x0, x1), cons(x2, x3)) -> equal_sort[a34](x0, x2) and equal_sort[a34](x1, x3)
   equal_sort[a34](cons(x0, x1), nil) -> false
   equal_sort[a34](nil, cons(x0, x1)) -> false
   equal_sort[a34](nil, nil) -> true
   equal_sort[a60](witness_sort[a60], witness_sort[a60]) -> true
   if2'(true, x122, y85, xs56) -> true
   if2'(false, x136, y95, cons(y37, xs25)) -> if2'(eq(x136, y37), x136, y37, xs25)
   if2'(false, x136, y95, nil) -> false
   del'(x150, nil) -> false
   equal_bool(eq(x54, y37), true) -> true | del'(x54, cons(y37, xs25)) -> true
   equal_bool(eq(x54, y37), true) -> false | del'(x54, cons(y37, xs25)) -> del'(x54, xs25)
   max(cons(x, nil)) -> x
   max(nil) -> 0
   equal_bool(ge(x12, y7), true) -> true | max(cons(x12, cons(y7, xs4))) -> max(cons(x12, xs4))
   equal_bool(ge(x12, y7), true) -> false | max(cons(x12, cons(y7, xs4))) -> max(cons(y7, xs4))
   eq(0, 0) -> true
   eq(0, s(y56)) -> false
   eq(s(x94), 0) -> false
   eq(s(x108), s(y75)) -> eq(x108, y75)
   if2(true, x122, y85, xs56) -> xs56
   if2(false, x136, y95, cons(y37, xs25)) -> cons(y95, if2(eq(x136, y37), x136, y37, xs25))
   if2(false, x136, y95, nil) -> cons(y95, nil)
   del(x150, nil) -> nil
   equal_bool(eq(x54, y37), true) -> true | del(x54, cons(y37, xs25)) -> xs25
   equal_bool(eq(x54, y37), true) -> false | del(x54, cons(y37, xs25)) -> cons(y37, del(x54, xs25))
   ge(0, 0) -> true
   ge(s(x177), 0) -> true
   ge(0, s(x191)) -> false
   ge(s(x205), s(y141)) -> ge(x205, y141)
   if1(true, x26, y17, nil) -> x26
   if1(true, x26, y17, cons(y7, xs4)) -> if1(ge(x26, y7), x26, y7, xs4)
   if1(false, x40, y27, nil) -> y27
   if1(false, x40, y27, cons(y7, xs4)) -> if1(ge(y27, y7), y27, y7, xs4)
   if1(true, x26, y17, xs11) -> 0
   if1(false, x40, y27, xs18) -> 0

using the following formula:
z0:sort[a34].(~(z0=nil)->del'(max(z0), z0)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation [EQUIVALENT]
                (10) YES
            (11) Formula
                (12) Conditional Evaluation [EQUIVALENT]
                (13) AND
                    (14) Formula
                        (15) Symbolic evaluation [EQUIVALENT]
                        (16) YES
                    (17) Formula
                        (18) Symbolic evaluation [EQUIVALENT]
                        (19) Formula
                        (20) Hypothesis Lifting [EQUIVALENT]
                        (21) Formula
                        (22) Symbolic evaluation under hypothesis [SOUND]
                        (23) Formula
                        (24) Hypothesis Lifting [EQUIVALENT]
                        (25) Formula
                        (26) Hypothesis Lifting [EQUIVALENT]
                        (27) Formula
                        (28) Conditional Evaluation [EQUIVALENT]
                        (29) AND
                            (30) Formula
                                (31) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (32) YES
                            (33) Formula
                                (34) Symbolic evaluation [EQUIVALENT]
                                (35) YES
    (36) Formula
        (37) Symbolic evaluation [EQUIVALENT]
        (38) YES
    (39) Formula
        (40) Symbolic evaluation [EQUIVALENT]
        (41) Formula
        (42) Conditional Evaluation [EQUIVALENT]
        (43) Formula
        (44) Conditional Evaluation [EQUIVALENT]
        (45) AND
            (46) Formula
                (47) Symbolic evaluation [EQUIVALENT]
                (48) YES
            (49) Formula
                (50) Conditional Evaluation [EQUIVALENT]
                (51) AND
                    (52) Formula
                        (53) Symbolic evaluation [EQUIVALENT]
                        (54) YES
                    (55) Formula
                        (56) Hypothesis Lifting [EQUIVALENT]
                        (57) Formula
                        (58) Conditional Evaluation [EQUIVALENT]
                        (59) Formula
                        (60) Symbolic evaluation [EQUIVALENT]
                        (61) YES
    (62) Formula
        (63) Symbolic evaluation [EQUIVALENT]
        (64) Formula
        (65) Conditional Evaluation [EQUIVALENT]
        (66) Formula
        (67) Conditional Evaluation [EQUIVALENT]
        (68) AND
            (69) Formula
                (70) Symbolic evaluation [EQUIVALENT]
                (71) YES
            (72) Formula
                (73) Symbolic evaluation under hypothesis [EQUIVALENT]
                (74) YES


----------------------------------------

(0)
Obligation:
Formula:
z0:sort[a34].(~(z0=nil)->del'(max(z0), z0)=true)

There are no hypotheses.




----------------------------------------

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm max(z0) generates the following cases:

1. Base Case:
Formula:
x:sort[a0].(~(cons(x, nil)=nil)->del'(max(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.





2. Base Case:
Formula:
(~(nil=nil)->del'(max(nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].(~(cons(x12, cons(y7, xs4))=nil)->del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true)

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true





2. Step Case:
Formula:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].(~(cons(x12, cons(y7, xs4))=nil)->del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true)

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
x:sort[a0].(~(cons(x, nil)=nil)->del'(max(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x:sort[a0].del'(x, cons(x, nil))=true
----------------------------------------

(5)
Obligation:
Formula:
x:sort[a0].del'(x, cons(x, nil))=true

There are no hypotheses.




----------------------------------------

(6) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
del'(0, cons(0, nil))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true






----------------------------------------

(7)
Complex Obligation (AND)

----------------------------------------

(8)
Obligation:
Formula:
del'(0, cons(0, nil))=true

There are no hypotheses.




----------------------------------------

(9) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(10)
YES

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a0].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true




----------------------------------------

(12) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=true





Formula:
n:sort[a0].del'(s(n), nil)=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=false






----------------------------------------

(13)
Complex Obligation (AND)

----------------------------------------

(14)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=true




----------------------------------------

(15) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(16)
YES

----------------------------------------

(17)
Obligation:
Formula:
n:sort[a0].del'(s(n), nil)=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=false




----------------------------------------

(18) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
False
----------------------------------------

(19)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=false




----------------------------------------

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].((del'(n, cons(n, nil))=true/\equal_bool(eq(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=false




----------------------------------------

(21)
Obligation:
Formula:
n:sort[a0].((del'(n, cons(n, nil))=true/\equal_bool(eq(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=false




----------------------------------------

(22) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_bool(eq(n, n), true)=false)

By using the following hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true

----------------------------------------

(23)
Obligation:
Formula:
n:sort[a0].~(equal_bool(eq(n, n), true)=false)

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_bool(eq(s(n), s(n)), true)=false




----------------------------------------

(24) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false))

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true




----------------------------------------

(25)
Obligation:
Formula:
n:sort[a0].(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false))

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true




----------------------------------------

(26) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(del'(n, cons(n, nil))=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

There are no hypotheses.




----------------------------------------

(27)
Obligation:
Formula:
n:sort[a0].(del'(n, cons(n, nil))=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

There are no hypotheses.




----------------------------------------

(28) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].(true=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(eq(n, n), true)=true





Formula:
n:sort[a0].(del'(n, nil)=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(eq(n, n), true)=false






----------------------------------------

(29)
Complex Obligation (AND)

----------------------------------------

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(eq(n, n), true)=true




----------------------------------------

(31) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a0].equal_bool(eq(n, n), true)=true

----------------------------------------

(32)
YES

----------------------------------------

(33)
Obligation:
Formula:
n:sort[a0].(del'(n, nil)=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(eq(n, n), true)=false




----------------------------------------

(34) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(35)
YES

----------------------------------------

(36)
Obligation:
Formula:
(~(nil=nil)->del'(max(nil), nil)=true)

There are no hypotheses.




----------------------------------------

(37) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(38)
YES

----------------------------------------

(39)
Obligation:
Formula:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].(~(cons(x12, cons(y7, xs4))=nil)->del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true)

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true




----------------------------------------

(40) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true
----------------------------------------

(41)
Obligation:
Formula:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true




----------------------------------------

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x12:sort[a0],xs4:sort[a34],y7:sort[a0].del'(max(cons(x12, xs4)), cons(x12, cons(y7, xs4)))=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true






----------------------------------------

(43)
Obligation:
Formula:
x12:sort[a0],xs4:sort[a34],y7:sort[a0].del'(max(cons(x12, xs4)), cons(x12, cons(y7, xs4)))=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true




----------------------------------------

(44) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=true





Formula:
x12:sort[a0],xs4:sort[a34],y7:sort[a0].del'(max(cons(x12, xs4)), cons(y7, xs4))=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false






----------------------------------------

(45)
Complex Obligation (AND)

----------------------------------------

(46)
Obligation:
Formula:
true=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=true




----------------------------------------

(47) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(48)
YES

----------------------------------------

(49)
Obligation:
Formula:
x12:sort[a0],xs4:sort[a34],y7:sort[a0].del'(max(cons(x12, xs4)), cons(y7, xs4))=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false




----------------------------------------

(50) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=true





Formula:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), xs4)=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=false






----------------------------------------

(51)
Complex Obligation (AND)

----------------------------------------

(52)
Obligation:
Formula:
true=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=true




----------------------------------------

(53) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(54)
YES

----------------------------------------

(55)
Obligation:
Formula:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), xs4)=true

Hypotheses:
x12:sort[a0],xs4:sort[a34].del'(max(cons(x12, xs4)), cons(x12, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=false




----------------------------------------

(56) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x12:sort[a0],xs4:sort[a34].(del'(max(cons(x12, xs4)), cons(x12, xs4))=true->del'(max(cons(x12, xs4)), xs4)=true)

Hypotheses:
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=false




----------------------------------------

(57)
Obligation:
Formula:
x12:sort[a0],xs4:sort[a34].(del'(max(cons(x12, xs4)), cons(x12, xs4))=true->del'(max(cons(x12, xs4)), xs4)=true)

Hypotheses:
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=false




----------------------------------------

(58) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x12:sort[a0],xs4:sort[a34].(del'(max(cons(x12, xs4)), xs4)=true->del'(max(cons(x12, xs4)), xs4)=true)

Hypotheses:
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=false






----------------------------------------

(59)
Obligation:
Formula:
x12:sort[a0],xs4:sort[a34].(del'(max(cons(x12, xs4)), xs4)=true->del'(max(cons(x12, xs4)), xs4)=true)

Hypotheses:
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=true
x12:sort[a0],xs4:sort[a34].equal_bool(eq(max(cons(x12, xs4)), x12), true)=false
x12:sort[a0],xs4:sort[a34],y7:sort[a0].equal_bool(eq(max(cons(x12, xs4)), y7), true)=false




----------------------------------------

(60) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(61)
YES

----------------------------------------

(62)
Obligation:
Formula:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].(~(cons(x12, cons(y7, xs4))=nil)->del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true)

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false




----------------------------------------

(63) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true
----------------------------------------

(64)
Obligation:
Formula:
x12:sort[a0],y7:sort[a0],xs4:sort[a34].del'(max(cons(x12, cons(y7, xs4))), cons(x12, cons(y7, xs4)))=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false




----------------------------------------

(65) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y7:sort[a0],xs4:sort[a34],x12:sort[a0].del'(max(cons(y7, xs4)), cons(x12, cons(y7, xs4)))=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false






----------------------------------------

(66)
Obligation:
Formula:
y7:sort[a0],xs4:sort[a34],x12:sort[a0].del'(max(cons(y7, xs4)), cons(x12, cons(y7, xs4)))=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false




----------------------------------------

(67) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false
y7:sort[a0],xs4:sort[a34],x12:sort[a0].equal_bool(eq(max(cons(y7, xs4)), x12), true)=true





Formula:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false
y7:sort[a0],xs4:sort[a34],x12:sort[a0].equal_bool(eq(max(cons(y7, xs4)), x12), true)=false






----------------------------------------

(68)
Complex Obligation (AND)

----------------------------------------

(69)
Obligation:
Formula:
true=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false
y7:sort[a0],xs4:sort[a34],x12:sort[a0].equal_bool(eq(max(cons(y7, xs4)), x12), true)=true




----------------------------------------

(70) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(71)
YES

----------------------------------------

(72)
Obligation:
Formula:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true

Hypotheses:
y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true
x12:sort[a0],y7:sort[a0].equal_bool(ge(x12, y7), true)=false
y7:sort[a0],xs4:sort[a34],x12:sort[a0].equal_bool(eq(max(cons(y7, xs4)), x12), true)=false




----------------------------------------

(73) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

y7:sort[a0],xs4:sort[a34].del'(max(cons(y7, xs4)), cons(y7, xs4))=true

----------------------------------------

(74)
YES

(43) Complex Obligation (AND)

(44) Obligation:

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

max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, cons(y, xs)) → if2(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)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) YES

(47) Obligation:

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

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

Q is empty.

(48) Overlay + Local Confluence (EQUIVALENT transformation)

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

(49) Obligation:

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

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(50) DependencyPairsProof (EQUIVALENT transformation)

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

(51) Obligation:

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

DEL'(x54, cons(y37, xs25)) → IF2'(eq(x54, y37), x54, y37, xs25)
DEL'(x54, cons(y37, xs25)) → EQ(x54, y37)
IF2'(false, x136, y95, xs63) → DEL'(x136, xs63)
MAX(cons(x12, cons(y7, xs4))) → IF1(ge(x12, y7), x12, y7, xs4)
MAX(cons(x12, cons(y7, xs4))) → GE(x12, y7)
IF1(true, x26, y17, xs11) → MAX(cons(x26, xs11))
IF1(false, x40, y27, xs18) → MAX(cons(y27, xs18))
DEL(x54, cons(y37, xs25)) → IF2(eq(x54, y37), x54, y37, xs25)
DEL(x54, cons(y37, xs25)) → EQ(x54, y37)
EQ(s(x108), s(y75)) → EQ(x108, y75)
IF2(false, x136, y95, xs63) → DEL(x136, xs63)
GE(s(x205), s(y141)) → GE(x205, y141)
EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → AND(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x0, x2)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x1, x3)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(52) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 4 less nodes.

(53) Complex Obligation (AND)

(54) Obligation:

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

EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x1, x3)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x0, x2)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(55) 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.

(56) Obligation:

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

EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x1, x3)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x0, x2)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(57) 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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(58) Obligation:

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

EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x1, x3)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x0, x2)

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

(59) 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[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x1, x3)
  The graph contains the following edges 1 > 1, 2 > 2

• EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x0, x2)
  The graph contains the following edges 1 > 1, 2 > 2

(60) YES

(61) Obligation:

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

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(62) 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.

(63) Obligation:

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

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(64) 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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(65) Obligation:

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

EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)

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

(66) 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[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
  The graph contains the following edges 1 > 1, 2 > 2

(67) YES

(68) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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:

GE(s(x205), s(y141)) → GE(x205, y141)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(72) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

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:

• GE(s(x205), s(y141)) → GE(x205, y141)
  The graph contains the following edges 1 > 1, 2 > 2

(74) YES

(75) Obligation:

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

EQ(s(x108), s(y75)) → EQ(x108, y75)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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:

EQ(s(x108), s(y75)) → EQ(x108, y75)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(79) Obligation:

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

EQ(s(x108), s(y75)) → EQ(x108, y75)

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:

• EQ(s(x108), s(y75)) → EQ(x108, y75)
  The graph contains the following edges 1 > 1, 2 > 2

(81) YES

(82) Obligation:

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

IF2(false, x136, y95, xs63) → DEL(x136, xs63)
DEL(x54, cons(y37, xs25)) → IF2(eq(x54, y37), x54, y37, xs25)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(83) 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.

(84) Obligation:

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

IF2(false, x136, y95, xs63) → DEL(x136, xs63)
DEL(x54, cons(y37, xs25)) → IF2(eq(x54, y37), x54, y37, xs25)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(85) 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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(86) Obligation:

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

IF2(false, x136, y95, xs63) → DEL(x136, xs63)
DEL(x54, cons(y37, xs25)) → IF2(eq(x54, y37), x54, y37, xs25)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)

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.

(87) 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(x54, cons(y37, xs25)) → IF2(eq(x54, y37), x54, y37, xs25)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2(false, x136, y95, xs63) → DEL(x136, xs63)
  The graph contains the following edges 2 >= 1, 4 >= 2

(88) YES

(89) Obligation:

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

IF1(true, x26, y17, xs11) → MAX(cons(x26, xs11))
MAX(cons(x12, cons(y7, xs4))) → IF1(ge(x12, y7), x12, y7, xs4)
IF1(false, x40, y27, xs18) → MAX(cons(y27, xs18))

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(90) 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.

(91) Obligation:

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

IF1(true, x26, y17, xs11) → MAX(cons(x26, xs11))
MAX(cons(x12, cons(y7, xs4))) → IF1(ge(x12, y7), x12, y7, xs4)
IF1(false, x40, y27, xs18) → MAX(cons(y27, xs18))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(92) 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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(93) Obligation:

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

IF1(true, x26, y17, xs11) → MAX(cons(x26, xs11))
MAX(cons(x12, cons(y7, xs4))) → IF1(ge(x12, y7), x12, y7, xs4)
IF1(false, x40, y27, xs18) → MAX(cons(y27, xs18))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(94) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF1(true, x26, y17, xs11) → MAX(cons(x26, xs11))
MAX(cons(x12, cons(y7, xs4))) → IF1(ge(x12, y7), x12, y7, xs4)
IF1(false, x40, y27, xs18) → MAX(cons(y27, xs18))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IF1(x₁, ..., x₄) ) = 2x₂ + 2x₃ + x₄ + 1

POL( ge(x₁, x₂) ) = 1

POL( 0 ) = 2

POL( true ) = 0

POL( s(x₁) ) = max{0, -1}

POL( false ) = 2

POL( MAX(x₁) ) = max{0, x₁ - 2}

POL( cons(x₁, x₂) ) = 2x₁ + x₂ + 2

The following usable rules [FROCOS05] were oriented: none

(95) Obligation:

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

ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(96) PisEmptyProof (EQUIVALENT transformation)

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

(97) YES

(98) Obligation:

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

IF2'(false, x136, y95, xs63) → DEL'(x136, xs63)
DEL'(x54, cons(y37, xs25)) → IF2'(eq(x54, y37), x54, y37, xs25)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
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[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](nil, nil) → true
equal_sort[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(99) 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.

(100) Obligation:

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

IF2'(false, x136, y95, xs63) → DEL'(x136, xs63)
DEL'(x54, cons(y37, xs25)) → IF2'(eq(x54, y37), x54, y37, xs25)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

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

(101) 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].

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(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[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](nil, nil)
equal_sort[a60](witness_sort[a60], witness_sort[a60])

(102) Obligation:

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

IF2'(false, x136, y95, xs63) → DEL'(x136, xs63)
DEL'(x54, cons(y37, xs25)) → IF2'(eq(x54, y37), x54, y37, xs25)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)

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.

(103) 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'(x54, cons(y37, xs25)) → IF2'(eq(x54, y37), x54, y37, xs25)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2'(false, x136, y95, xs63) → DEL'(x136, xs63)
  The graph contains the following edges 2 >= 1, 4 >= 2

(104) YES