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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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(h(del(max(cons(x, xs)), cons(x, xs))))
SORT(cons(x, xs)) → H(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)
H(cons(x, xs)) → H(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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

H(cons(x, xs)) → H(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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

H(cons(x, xs)) → H(xs)

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))
h(nil)
h(cons(x0, x1))

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))
h(nil)
h(cons(x0, x1))

(11) Obligation:

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

H(cons(x, xs)) → H(xs)

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:

• H(cons(x, xs)) → H(xs)
  The graph contains the following edges 1 > 1

(13) YES

(14) 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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

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))
h(nil)
h(cons(x0, x1))

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))
h(nil)
h(cons(x0, x1))

(18) 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.

(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:

• GE(s(x), s(y)) → GE(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:

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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

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))
h(nil)
h(cons(x0, x1))

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))
h(nil)
h(cons(x0, x1))

(25) 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.

(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:

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

(27) YES

(28) 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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

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))
h(nil)
h(cons(x0, x1))

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))
h(nil)
h(cons(x0, x1))

(32) 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.

(33) 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

(34) YES

(35) 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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(36) 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.

(37) 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))
h(nil)
h(cons(x0, x1))

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

(38) 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))
h(nil)
h(cons(x0, x1))

(39) 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.

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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 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

(41) Obligation:

Q DP problem:
P is empty.
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.

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) YES

(44) Obligation:

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

SORT(cons(x, xs)) → SORT(h(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(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

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))
h(nil)
h(cons(x0, x1))

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

(45) 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.

(46) Obligation:

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

SORT(cons(x, xs)) → SORT(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(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))
h(nil)
h(cons(x0, x1))

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

(47) 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))

(48) Obligation:

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

SORT(cons(x, xs)) → SORT(h(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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(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))
h(nil)
h(cons(x0, x1))

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

(49) Induction-Processor (SOUND transformation)

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


This order was computed:
Polynomial interpretation [POLO]:

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

At least one of these decreasing rules is always used after the deleted DP:
if2(true, x1609, y1039, xs809) → xs809


The following formula is valid:
z0:sort[a36].(¬(z0 =nil)→del'(max(z0 ), z0 )=true)


The transformed set:
del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → true


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

   [x, x0, x1, x2, x3, x160, y103, xs80, x175, y113, y37, xs29, x190, x58, x13, y7, xs5, x87, xs44, y74, x130, x145, y93, x219, x234, x249, y159, x28, y17, x43, y27, xs13, xs21]
   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[a36](cons(x0, x1), cons(x2, x3)) -> equal_sort[a36](x0, x2) and equal_sort[a36](x1, x3)
   equal_sort[a36](cons(x0, x1), nil) -> false
   equal_sort[a36](nil, cons(x0, x1)) -> false
   equal_sort[a36](nil, nil) -> true
   equal_sort[a64](witness_sort[a64], witness_sort[a64]) -> true
   if2'(true, x160, y103, xs80) -> true
   if2'(false, x175, y113, cons(y37, xs29)) -> if2'(eq(x175, y37), x175, y37, xs29)
   if2'(false, x175, y113, nil) -> false
   del'(x190, nil) -> false
   equal_bool(eq(x58, y37), true) -> true | del'(x58, cons(y37, xs29)) -> true
   equal_bool(eq(x58, y37), true) -> false | del'(x58, cons(y37, xs29)) -> del'(x58, xs29)
   max(cons(x, nil)) -> x
   max(nil) -> 0
   equal_bool(ge(x13, y7), true) -> true | max(cons(x13, cons(y7, xs5))) -> max(cons(x13, xs5))
   equal_bool(ge(x13, y7), true) -> false | max(cons(x13, cons(y7, xs5))) -> max(cons(y7, xs5))
   h(nil) -> nil
   h(cons(x87, xs44)) -> cons(x87, h(xs44))
   eq(0, 0) -> true
   eq(0, s(y74)) -> false
   eq(s(x130), 0) -> false
   eq(s(x145), s(y93)) -> eq(x145, y93)
   if2(true, x160, y103, xs80) -> xs80
   if2(false, x175, y113, cons(y37, xs29)) -> cons(y113, if2(eq(x175, y37), x175, y37, xs29))
   if2(false, x175, y113, nil) -> cons(y113, nil)
   del(x190, nil) -> nil
   equal_bool(eq(x58, y37), true) -> true | del(x58, cons(y37, xs29)) -> xs29
   equal_bool(eq(x58, y37), true) -> false | del(x58, cons(y37, xs29)) -> cons(y37, del(x58, xs29))
   ge(0, 0) -> true
   ge(s(x219), 0) -> true
   ge(0, s(x234)) -> false
   ge(s(x249), s(y159)) -> ge(x249, y159)
   if1(true, x28, y17, nil) -> x28
   if1(true, x28, y17, cons(y7, xs5)) -> if1(ge(x28, y7), x28, y7, xs5)
   if1(false, x43, y27, nil) -> y27
   if1(false, x43, y27, cons(y7, xs5)) -> if1(ge(y27, y7), y27, y7, xs5)
   if1(true, x28, y17, xs13) -> 0
   if1(false, x43, y27, xs21) -> 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, x160, y103, xs80, x175, y113, y37, xs29, x190, x58, x13, y7, xs5, x87, xs44, y74, x130, x145, y93, x219, x234, x249, y159, x28, y17, x43, y27, xs13, xs21]
   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[a36](cons(x0, x1), cons(x2, x3)) -> equal_sort[a36](x0, x2) and equal_sort[a36](x1, x3)
   equal_sort[a36](cons(x0, x1), nil) -> false
   equal_sort[a36](nil, cons(x0, x1)) -> false
   equal_sort[a36](nil, nil) -> true
   equal_sort[a64](witness_sort[a64], witness_sort[a64]) -> true
   if2'(true, x160, y103, xs80) -> true
   if2'(false, x175, y113, cons(y37, xs29)) -> if2'(eq(x175, y37), x175, y37, xs29)
   if2'(false, x175, y113, nil) -> false
   del'(x190, nil) -> false
   equal_bool(eq(x58, y37), true) -> true | del'(x58, cons(y37, xs29)) -> true
   equal_bool(eq(x58, y37), true) -> false | del'(x58, cons(y37, xs29)) -> del'(x58, xs29)
   max(cons(x, nil)) -> x
   max(nil) -> 0
   equal_bool(ge(x13, y7), true) -> true | max(cons(x13, cons(y7, xs5))) -> max(cons(x13, xs5))
   equal_bool(ge(x13, y7), true) -> false | max(cons(x13, cons(y7, xs5))) -> max(cons(y7, xs5))
   h(nil) -> nil
   h(cons(x87, xs44)) -> cons(x87, h(xs44))
   eq(0, 0) -> true
   eq(0, s(y74)) -> false
   eq(s(x130), 0) -> false
   eq(s(x145), s(y93)) -> eq(x145, y93)
   if2(true, x160, y103, xs80) -> xs80
   if2(false, x175, y113, cons(y37, xs29)) -> cons(y113, if2(eq(x175, y37), x175, y37, xs29))
   if2(false, x175, y113, nil) -> cons(y113, nil)
   del(x190, nil) -> nil
   equal_bool(eq(x58, y37), true) -> true | del(x58, cons(y37, xs29)) -> xs29
   equal_bool(eq(x58, y37), true) -> false | del(x58, cons(y37, xs29)) -> cons(y37, del(x58, xs29))
   ge(0, 0) -> true
   ge(s(x219), 0) -> true
   ge(0, s(x234)) -> false
   ge(s(x249), s(y159)) -> ge(x249, y159)
   if1(true, x28, y17, nil) -> x28
   if1(true, x28, y17, cons(y7, xs5)) -> if1(ge(x28, y7), x28, y7, xs5)
   if1(false, x43, y27, nil) -> y27
   if1(false, x43, y27, cons(y7, xs5)) -> if1(ge(y27, y7), y27, y7, xs5)
   if1(true, x28, y17, xs13) -> 0
   if1(false, x43, y27, xs21) -> 0

using the following formula:
z0:sort[a36].(~(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[a36].(~(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:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].(~(cons(x13, cons(y7, xs5))=nil)->del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true)

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true





2. Step Case:
Formula:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].(~(cons(x13, cons(y7, xs5))=nil)->del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true)

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, 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:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].(~(cons(x13, cons(y7, xs5))=nil)->del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true)

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true




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

(40) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true
----------------------------------------

(41)
Obligation:
Formula:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true




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

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x13:sort[a0],xs5:sort[a36],y7:sort[a0].del'(max(cons(x13, xs5)), cons(x13, cons(y7, xs5)))=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true






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

(43)
Obligation:
Formula:
x13:sort[a0],xs5:sort[a36],y7:sort[a0].del'(max(cons(x13, xs5)), cons(x13, cons(y7, xs5)))=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true




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

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

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=true





Formula:
x13:sort[a0],xs5:sort[a36],y7:sort[a0].del'(max(cons(x13, xs5)), cons(y7, xs5))=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false






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

(45)
Complex Obligation (AND)

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

(46)
Obligation:
Formula:
true=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=true




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

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

(48)
YES

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

(49)
Obligation:
Formula:
x13:sort[a0],xs5:sort[a36],y7:sort[a0].del'(max(cons(x13, xs5)), cons(y7, xs5))=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false




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

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

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), y7), true)=true





Formula:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), xs5)=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), y7), true)=false






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

(51)
Complex Obligation (AND)

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

(52)
Obligation:
Formula:
true=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), 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:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), xs5)=true

Hypotheses:
x13:sort[a0],xs5:sort[a36].del'(max(cons(x13, xs5)), cons(x13, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), y7), true)=false




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

(56) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x13:sort[a0],xs5:sort[a36].(del'(max(cons(x13, xs5)), cons(x13, xs5))=true->del'(max(cons(x13, xs5)), xs5)=true)

Hypotheses:
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), y7), true)=false




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

(57)
Obligation:
Formula:
x13:sort[a0],xs5:sort[a36].(del'(max(cons(x13, xs5)), cons(x13, xs5))=true->del'(max(cons(x13, xs5)), xs5)=true)

Hypotheses:
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), y7), true)=false




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

(58) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x13:sort[a0],xs5:sort[a36].(del'(max(cons(x13, xs5)), xs5)=true->del'(max(cons(x13, xs5)), xs5)=true)

Hypotheses:
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), y7), true)=false






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

(59)
Obligation:
Formula:
x13:sort[a0],xs5:sort[a36].(del'(max(cons(x13, xs5)), xs5)=true->del'(max(cons(x13, xs5)), xs5)=true)

Hypotheses:
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=true
x13:sort[a0],xs5:sort[a36].equal_bool(eq(max(cons(x13, xs5)), x13), true)=false
x13:sort[a0],xs5:sort[a36],y7:sort[a0].equal_bool(eq(max(cons(x13, xs5)), 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:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].(~(cons(x13, cons(y7, xs5))=nil)->del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true)

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false




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

(63) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true
----------------------------------------

(64)
Obligation:
Formula:
x13:sort[a0],y7:sort[a0],xs5:sort[a36].del'(max(cons(x13, cons(y7, xs5))), cons(x13, cons(y7, xs5)))=true

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false




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

(65) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y7:sort[a0],xs5:sort[a36],x13:sort[a0].del'(max(cons(y7, xs5)), cons(x13, cons(y7, xs5)))=true

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false






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

(66)
Obligation:
Formula:
y7:sort[a0],xs5:sort[a36],x13:sort[a0].del'(max(cons(y7, xs5)), cons(x13, cons(y7, xs5)))=true

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, 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],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false
y7:sort[a0],xs5:sort[a36],x13:sort[a0].equal_bool(eq(max(cons(y7, xs5)), x13), true)=true





Formula:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false
y7:sort[a0],xs5:sort[a36],x13:sort[a0].equal_bool(eq(max(cons(y7, xs5)), x13), true)=false






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

(68)
Complex Obligation (AND)

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

(69)
Obligation:
Formula:
true=true

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false
y7:sort[a0],xs5:sort[a36],x13:sort[a0].equal_bool(eq(max(cons(y7, xs5)), x13), 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],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true

Hypotheses:
y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true
x13:sort[a0],y7:sort[a0].equal_bool(ge(x13, y7), true)=false
y7:sort[a0],xs5:sort[a36],x13:sort[a0].equal_bool(eq(max(cons(y7, xs5)), x13), true)=false




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

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

y7:sort[a0],xs5:sort[a36].del'(max(cons(y7, xs5)), cons(y7, xs5))=true

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

(74)
YES

(50) Complex Obligation (AND)

(51) 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)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(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))
h(nil)
h(cons(x0, x1))

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

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) YES

(54) Obligation:

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

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → true

Q is empty.

(55) Overlay + Local Confluence (EQUIVALENT transformation)

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

(56) Obligation:

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

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(57) DependencyPairsProof (EQUIVALENT transformation)

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

(58) Obligation:

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

DEL'(x58, cons(y37, xs29)) → IF2'(eq(x58, y37), x58, y37, xs29)
DEL'(x58, cons(y37, xs29)) → EQ(x58, y37)
IF2'(false, x175, y113, xs88) → DEL'(x175, xs88)
MAX(cons(x13, cons(y7, xs5))) → IF1(ge(x13, y7), x13, y7, xs5)
MAX(cons(x13, cons(y7, xs5))) → GE(x13, y7)
IF1(true, x28, y17, xs13) → MAX(cons(x28, xs13))
IF1(false, x43, y27, xs21) → MAX(cons(y27, xs21))
DEL(x58, cons(y37, xs29)) → IF2(eq(x58, y37), x58, y37, xs29)
DEL(x58, cons(y37, xs29)) → EQ(x58, y37)
H(cons(x87, xs44)) → H(xs44)
EQ(s(x145), s(y93)) → EQ(x145, y93)
IF2(false, x175, y113, xs88) → DEL(x175, xs88)
GE(s(x249), s(y159)) → GE(x249, y159)
EQUAL_SORT[A0](s(x0), s(x1)) → EQUAL_SORT[A0](x0, x1)
EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → AND(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x0, x2)
EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x1, x3)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(59) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 4 less nodes.

(60) Complex Obligation (AND)

(61) Obligation:

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

EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x1, x3)
EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x0, x2)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x1, x3)
EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(65) Obligation:

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

EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x1, x3)
EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x0, x2)

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[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x1, x3)
  The graph contains the following edges 1 > 1, 2 > 2

• EQUAL_SORT[A36](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A36](x0, x2)
  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:

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

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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:

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(72) 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.

(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:

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

GE(s(x249), s(y159)) → GE(x249, y159)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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:

GE(s(x249), s(y159)) → GE(x249, y159)

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(79) Obligation:

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

GE(s(x249), s(y159)) → GE(x249, y159)

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:

• GE(s(x249), s(y159)) → GE(x249, y159)
  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:

EQ(s(x145), s(y93)) → EQ(x145, y93)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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:

EQ(s(x145), s(y93)) → EQ(x145, y93)

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(86) Obligation:

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

EQ(s(x145), s(y93)) → EQ(x145, y93)

R is empty.
Q is empty.
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:

• EQ(s(x145), s(y93)) → EQ(x145, y93)
  The graph contains the following edges 1 > 1, 2 > 2

(88) YES

(89) Obligation:

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

H(cons(x87, xs44)) → H(xs44)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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:

H(cons(x87, xs44)) → H(xs44)

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(93) Obligation:

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

H(cons(x87, xs44)) → H(xs44)

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

(94) 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:

• H(cons(x87, xs44)) → H(xs44)
  The graph contains the following edges 1 > 1

(95) YES

(96) Obligation:

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

IF2(false, x175, y113, xs88) → DEL(x175, xs88)
DEL(x58, cons(y37, xs29)) → IF2(eq(x58, y37), x58, y37, xs29)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(97) 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.

(98) Obligation:

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

IF2(false, x175, y113, xs88) → DEL(x175, xs88)
DEL(x58, cons(y37, xs29)) → IF2(eq(x58, y37), x58, y37, xs29)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(99) 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))
h(nil)
h(cons(x0, 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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(100) Obligation:

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

IF2(false, x175, y113, xs88) → DEL(x175, xs88)
DEL(x58, cons(y37, xs29)) → IF2(eq(x58, y37), x58, y37, xs29)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)

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.

(101) 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(x58, cons(y37, xs29)) → IF2(eq(x58, y37), x58, y37, xs29)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2(false, x175, y113, xs88) → DEL(x175, xs88)
  The graph contains the following edges 2 >= 1, 4 >= 2

(102) YES

(103) Obligation:

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

IF1(true, x28, y17, xs13) → MAX(cons(x28, xs13))
MAX(cons(x13, cons(y7, xs5))) → IF1(ge(x13, y7), x13, y7, xs5)
IF1(false, x43, y27, xs21) → MAX(cons(y27, xs21))

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(104) 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.

(105) Obligation:

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

IF1(true, x28, y17, xs13) → MAX(cons(x28, xs13))
MAX(cons(x13, cons(y7, xs5))) → IF1(ge(x13, y7), x13, y7, xs5)
IF1(false, x43, y27, xs21) → MAX(cons(y27, xs21))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(106) 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(107) Obligation:

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

IF1(true, x28, y17, xs13) → MAX(cons(x28, xs13))
MAX(cons(x13, cons(y7, xs5))) → IF1(ge(x13, y7), x13, y7, xs5)
IF1(false, x43, y27, xs21) → MAX(cons(y27, xs21))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)

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.

(108) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MAX(cons(x13, cons(y7, xs5))) → IF1(ge(x13, y7), x13, y7, xs5)

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

POL(IF1(x1, x2, x3, x4)) =

1

+

[ 0, 0 ]

·

x1

+

[ 0, 0 ]

·

x2

+

[ 0, 0 ]

·

x3

+

[ 1, 0 ]

·

x4

POL(true) =

/ 0 \ \ 0 /

POL(MAX(x1)) =

0

+

[ 1, 0 ]

·

x1

POL(cons(x1, x2)) =

/ 1 \ \ 0 /

+

/ 0 0 \ \ 0 0 /

·

x1

+

/ 1 0 \ \ 0 0 /

·

x2

POL(ge(x1, x2)) =

/ 0 \ \ 0 /

+

/ 0 0 \ \ 0 0 /

·

x1

+

/ 0 1 \ \ 0 0 /

·

x2

POL(false) =

/ 1 \ \ 0 /

POL(0) =

/ 0 \ \ 1 /

POL(s(x1)) =

/ 0 \ \ 0 /

+

/ 0 0 \ \ 1 0 /

·

x1

The following usable rules [FROCOS05] were oriented: none

(109) Obligation:

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

IF1(true, x28, y17, xs13) → MAX(cons(x28, xs13))
IF1(false, x43, y27, xs21) → MAX(cons(y27, xs21))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)

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.

(110) DependencyGraphProof (EQUIVALENT transformation)

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

(111) TRUE

(112) Obligation:

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

IF2'(false, x175, y113, xs88) → DEL'(x175, xs88)
DEL'(x58, cons(y37, xs29)) → IF2'(eq(x58, y37), x58, y37, xs29)

The TRS R consists of the following rules:

del'(x58, cons(y37, xs29)) → if2'(eq(x58, y37), x58, y37, xs29)
if2'(true, x160, y103, xs80) → true
if2'(false, x175, y113, xs88) → del'(x175, xs88)
del'(x190, nil) → false
max(cons(x, nil)) → x
max(cons(x13, cons(y7, xs5))) → if1(ge(x13, y7), x13, y7, xs5)
if1(true, x28, y17, xs13) → max(cons(x28, xs13))
if1(false, x43, y27, xs21) → max(cons(y27, xs21))
del(x58, cons(y37, xs29)) → if2(eq(x58, y37), x58, y37, xs29)
h(nil) → nil
h(cons(x87, xs44)) → cons(x87, h(xs44))
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)
if2(true, x160, y103, xs80) → xs80
if2(false, x175, y113, xs88) → cons(y113, del(x175, xs88))
del(x190, nil) → nil
ge(0, 0) → true
ge(s(x219), 0) → true
ge(0, s(x234)) → false
ge(s(x249), s(y159)) → ge(x249, y159)
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → 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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(113) 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.

(114) Obligation:

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

IF2'(false, x175, y113, xs88) → DEL'(x175, xs88)
DEL'(x58, cons(y37, xs29)) → IF2'(eq(x58, y37), x58, y37, xs29)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)

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))
h(nil)
h(cons(x0, x1))
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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

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

(115) 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))
h(nil)
h(cons(x0, 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[a36](cons(x0, x1), cons(x2, x3))
equal_sort[a36](cons(x0, x1), nil)
equal_sort[a36](nil, cons(x0, x1))
equal_sort[a36](nil, nil)
equal_sort[a64](witness_sort[a64], witness_sort[a64])

(116) Obligation:

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

IF2'(false, x175, y113, xs88) → DEL'(x175, xs88)
DEL'(x58, cons(y37, xs29)) → IF2'(eq(x58, y37), x58, y37, xs29)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x130), 0) → false
eq(s(x145), s(y93)) → eq(x145, y93)

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.

(117) 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'(x58, cons(y37, xs29)) → IF2'(eq(x58, y37), x58, y37, xs29)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2'(false, x175, y113, xs88) → DEL'(x175, xs88)
  The graph contains the following edges 2 >= 1, 4 >= 2

(118) YES