Termination w.r.t. Q proof of /tmp/tmphq5dGw/maxsortcondition.trs

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔)

        ↳9 QDP

        ↳10 QReductionProof (⇔)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔)

        ↳16 QDP

        ↳17 QReductionProof (⇔)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔)

        ↳23 QDP

        ↳24 QReductionProof (⇔)

        ↳25 QDP

        ↳26 QDPSizeChangeProof (⇔)

        ↳27 YES

    ↳28 QDP

        ↳29 UsableRulesProof (⇔)

        ↳30 QDP

        ↳31 QReductionProof (⇔)

        ↳32 QDP

        ↳33 QDPOrderProof (⇔)

        ↳34 QDP

        ↳35 DependencyGraphProof (⇔)

        ↳36 TRUE

    ↳37 QDP

        ↳38 UsableRulesProof (⇔)

        ↳39 QDP

        ↳40 QReductionProof (⇔)

        ↳41 QDP

        ↳42 Narrowing (⇔)

        ↳43 QDP

        ↳44 DependencyGraphProof (⇔)

        ↳45 QDP

        ↳46 UsableRulesProof (⇔)

        ↳47 QDP

        ↳48 QReductionProof (⇔)

        ↳49 QDP

        ↳50 Instantiation (⇔)

        ↳51 QDP

        ↳52 Induction-Processor (⇐)

        ↳53 AND

            ↳54 QDP

                ↳55 DependencyGraphProof (⇔)

                ↳56 TRUE

            ↳57 QTRS

                ↳58 Overlay + Local Confluence (⇔)

                ↳59 QTRS

                ↳60 DependencyPairsProof (⇔)

                ↳61 QDP

                ↳62 DependencyGraphProof (⇔)

                ↳63 AND

                    ↳64 QDP

                        ↳65 UsableRulesProof (⇔)

                        ↳66 QDP

                        ↳67 QReductionProof (⇔)

                        ↳68 QDP

                        ↳69 QDPSizeChangeProof (⇔)

                        ↳70 YES

                    ↳71 QDP

                        ↳72 UsableRulesProof (⇔)

                        ↳73 QDP

                        ↳74 QReductionProof (⇔)

                        ↳75 QDP

                        ↳76 QDPSizeChangeProof (⇔)

                        ↳77 YES

                    ↳78 QDP

                        ↳79 UsableRulesProof (⇔)

                        ↳80 QDP

                        ↳81 QReductionProof (⇔)

                        ↳82 QDP

                        ↳83 QDPSizeChangeProof (⇔)

                        ↳84 YES

                    ↳85 QDP

                        ↳86 UsableRulesProof (⇔)

                        ↳87 QDP

                        ↳88 QReductionProof (⇔)

                        ↳89 QDP

                        ↳90 QDPSizeChangeProof (⇔)

                        ↳91 YES

                    ↳92 QDP

                        ↳93 UsableRulesProof (⇔)

                        ↳94 QDP

                        ↳95 QReductionProof (⇔)

                        ↳96 QDP

                        ↳97 QDPSizeChangeProof (⇔)

                        ↳98 YES

                    ↳99 QDP

                        ↳100 UsableRulesProof (⇔)

                        ↳101 QDP

                        ↳102 QReductionProof (⇔)

                        ↳103 QDP

                        ↳104 QDPOrderProof (⇔)

                        ↳105 QDP

                        ↳106 DependencyGraphProof (⇔)

                        ↳107 TRUE

                    ↳108 QDP

                        ↳109 UsableRulesProof (⇔)

                        ↳110 QDP

                        ↳111 QReductionProof (⇔)

                        ↳112 QDP

                        ↳113 QDPSizeChangeProof (⇔)

                        ↳114 YES

(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(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
MAX(cons(x, cons(y, xs))) → GE(x, y)
IF1(true, x, y, xs) → MAX(cons(x, xs))
IF1(false, x, y, xs) → MAX(cons(y, xs))
DEL(x, cons(y, xs)) → IF2(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF2(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
SORT(xs) → IF3(empty(xs), xs)
SORT(xs) → EMPTY(xs)
IF3(false, xs) → SORT(del(max(xs), xs))
IF3(false, xs) → DEL(max(xs), xs)
IF3(false, xs) → MAX(xs)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(10) QReductionProof (EQUIVALENT transformation)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(11) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(13) YES

(14) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(17) QReductionProof (EQUIVALENT transformation)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(18) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(20) YES

(21) Obligation:

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

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

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(22) UsableRulesProof (EQUIVALENT transformation)

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

(23) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(24) QReductionProof (EQUIVALENT transformation)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(25) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(27) YES

(28) Obligation:

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

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

The TRS R consists of the following rules:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(29) UsableRulesProof (EQUIVALENT transformation)

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

(30) Obligation:

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

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

The TRS R consists of the following rules:

ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(31) QReductionProof (EQUIVALENT transformation)

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

max(nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, nil)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
sort(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))

(32) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

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

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(xs) → IF3(empty(xs), 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(xs) → if3(empty(xs), xs)
if3(true, xs) → nil
if3(false, xs) → sort(del(max(xs), xs))
empty(nil) → true
empty(cons(x, xs)) → false
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(38) UsableRulesProof (EQUIVALENT transformation)

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

(39) Obligation:

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

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(xs) → IF3(empty(xs), xs)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
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)
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))
ge(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(x0)
if3(true, x0)
if3(false, x0)
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(40) QReductionProof (EQUIVALENT transformation)

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

sort(x0)
if3(true, x0)
if3(false, x0)

(41) Obligation:

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

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(xs) → IF3(empty(xs), xs)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
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)
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))
ge(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))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(42) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SORT(xs) → IF3(empty(xs), xs) at position [0] we obtained the following new rules [LPAR04]:

SORT(nil) → IF3(true, nil)
SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))

(43) Obligation:

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

IF3(false, xs) → SORT(del(max(xs), xs))
SORT(nil) → IF3(true, nil)
SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
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)
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))
ge(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))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(44) DependencyGraphProof (EQUIVALENT transformation)

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

(45) Obligation:

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

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, xs) → SORT(del(max(xs), xs))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, xs)) → false
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)
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))
ge(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))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(46) 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.

(47) Obligation:

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

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, xs) → SORT(del(max(xs), 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)
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))
ge(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))
empty(nil)
empty(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(48) 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].

empty(nil)
empty(cons(x0, x1))

(49) Obligation:

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

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, xs) → SORT(del(max(xs), 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)
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))
ge(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(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(50) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(false, xs) → SORT(del(max(xs), xs)) we obtained the following new rules [LPAR04]:

IF3(false, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1)))

(51) Obligation:

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

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))
IF3(false, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1)))

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

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

(52) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
IF3(false, cons(z0, z1)) → SORT(del(max(cons(z0, z1)), cons(z0, z1)))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1
POL(IF3(x₁, x₂)) = x₂
POL(SORT(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(del(x₁, x₂)) = x₂
POL(eq(x₁, x₂)) = x₂
POL(false) = 1
POL(ge(x₁, x₂)) = 1
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₂ + x₃ + x₄
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(max(x₁)) = 1 + x₁
POL(nil) = 1
POL(s(x₁)) = 1 + x₁
POL(true) = 1

At least one of these decreasing rules is always used after the deleted DP:
if2(true, x1499, y1039, xs689) → xs689

The following formula is valid:
z2:sort[a34].(¬(z2 =nil)→del'(max(z2 ), z2 )=true)

The transformed set:
del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

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

   [x, x0, x1, x2, x3, x67, x81, y55, xs37, x149, y103, xs68, x163, y113, x11, x25, y16, xs10, y74, x121, x135, y93, x177, x191, x205, y141, x39, y26, x53, y36, xs17, xs24]
   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[a33](0, 0) -> true
   equal_sort[a33](0, s(x0)) -> false
   equal_sort[a33](s(x0), 0) -> false
   equal_sort[a33](s(x0), s(x1)) -> equal_sort[a33](x0, x1)
   equal_sort[a34](nil, nil) -> true
   equal_sort[a34](nil, cons(x0, x1)) -> false
   equal_sort[a34](cons(x0, x1), nil) -> false
   equal_sort[a34](cons(x0, x1), cons(x2, x3)) -> equal_sort[a34](x0, x2) and equal_sort[a34](x1, x3)
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true
   del'(x67, nil) -> false
   equal_bool(eq(x81, y55), true) -> true | del'(x81, cons(y55, xs37)) -> true
   equal_bool(eq(x81, y55), true) -> false | del'(x81, cons(y55, xs37)) -> del'(x81, xs37)
   if2'(true, x149, y103, xs68) -> true
   if2'(false, x163, y113, nil) -> false
   if2'(false, x163, y113, cons(y55, xs37)) -> if2'(eq(x163, y55), x163, y55, xs37)
   max(nil) -> 0
   max(cons(x11, nil)) -> x11
   equal_bool(ge(x25, y16), true) -> true | max(cons(x25, cons(y16, xs10))) -> max(cons(x25, xs10))
   equal_bool(ge(x25, y16), true) -> false | max(cons(x25, cons(y16, xs10))) -> max(cons(y16, xs10))
   del(x67, nil) -> nil
   equal_bool(eq(x81, y55), true) -> true | del(x81, cons(y55, xs37)) -> xs37
   equal_bool(eq(x81, y55), true) -> false | del(x81, cons(y55, xs37)) -> cons(y55, del(x81, xs37))
   eq(0, 0) -> true
   eq(0, s(y74)) -> false
   eq(s(x121), 0) -> false
   eq(s(x135), s(y93)) -> eq(x135, y93)
   if2(true, x149, y103, xs68) -> xs68
   if2(false, x163, y113, nil) -> cons(y113, nil)
   if2(false, x163, y113, cons(y55, xs37)) -> cons(y113, if2(eq(x163, y55), x163, y55, xs37))
   ge(x177, 0) -> true
   ge(0, s(x191)) -> false
   ge(s(x205), s(y141)) -> ge(x205, y141)
   if1(true, x39, y26, nil) -> x39
   if1(true, x39, y26, cons(y16, xs10)) -> if1(ge(x39, y16), x39, y16, xs10)
   if1(false, x53, y36, nil) -> y36
   if1(false, x53, y36, cons(y16, xs10)) -> if1(ge(y36, y16), y36, y16, xs10)
   if1(true, x39, y26, xs17) -> 0
   if1(false, x53, y36, xs24) -> 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, x67, x81, y55, xs37, x149, y103, xs68, x163, y113, x11, x25, y16, xs10, y74, x121, x135, y93, x177, x191, x205, y141, x39, y26, x53, y36, xs17, xs24]
   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[a33](0, 0) -> true
   equal_sort[a33](0, s(x0)) -> false
   equal_sort[a33](s(x0), 0) -> false
   equal_sort[a33](s(x0), s(x1)) -> equal_sort[a33](x0, x1)
   equal_sort[a34](nil, nil) -> true
   equal_sort[a34](nil, cons(x0, x1)) -> false
   equal_sort[a34](cons(x0, x1), nil) -> false
   equal_sort[a34](cons(x0, x1), cons(x2, x3)) -> equal_sort[a34](x0, x2) and equal_sort[a34](x1, x3)
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true
   del'(x67, nil) -> false
   equal_bool(eq(x81, y55), true) -> true | del'(x81, cons(y55, xs37)) -> true
   equal_bool(eq(x81, y55), true) -> false | del'(x81, cons(y55, xs37)) -> del'(x81, xs37)
   if2'(true, x149, y103, xs68) -> true
   if2'(false, x163, y113, nil) -> false
   if2'(false, x163, y113, cons(y55, xs37)) -> if2'(eq(x163, y55), x163, y55, xs37)
   max(nil) -> 0
   max(cons(x11, nil)) -> x11
   equal_bool(ge(x25, y16), true) -> true | max(cons(x25, cons(y16, xs10))) -> max(cons(x25, xs10))
   equal_bool(ge(x25, y16), true) -> false | max(cons(x25, cons(y16, xs10))) -> max(cons(y16, xs10))
   del(x67, nil) -> nil
   equal_bool(eq(x81, y55), true) -> true | del(x81, cons(y55, xs37)) -> xs37
   equal_bool(eq(x81, y55), true) -> false | del(x81, cons(y55, xs37)) -> cons(y55, del(x81, xs37))
   eq(0, 0) -> true
   eq(0, s(y74)) -> false
   eq(s(x121), 0) -> false
   eq(s(x135), s(y93)) -> eq(x135, y93)
   if2(true, x149, y103, xs68) -> xs68
   if2(false, x163, y113, nil) -> cons(y113, nil)
   if2(false, x163, y113, cons(y55, xs37)) -> cons(y113, if2(eq(x163, y55), x163, y55, xs37))
   ge(x177, 0) -> true
   ge(0, s(x191)) -> false
   ge(s(x205), s(y141)) -> ge(x205, y141)
   if1(true, x39, y26, nil) -> x39
   if1(true, x39, y26, cons(y16, xs10)) -> if1(ge(x39, y16), x39, y16, xs10)
   if1(false, x53, y36, nil) -> y36
   if1(false, x53, y36, cons(y16, xs10)) -> if1(ge(y36, y16), y36, y16, xs10)
   if1(true, x39, y26, xs17) -> 0
   if1(false, x53, y36, xs24) -> 0

using the following formula:
z2:sort[a34].(~(z2=nil)->del'(max(z2), z2)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT]
        (8) Formula
        (9) Induction by data structure [EQUIVALENT]
        (10) AND
            (11) Formula
                (12) Symbolic evaluation [EQUIVALENT]
                (13) YES
            (14) Formula
                (15) Conditional Evaluation [EQUIVALENT]
                (16) AND
                    (17) Formula
                        (18) Symbolic evaluation [EQUIVALENT]
                        (19) YES
                    (20) Formula
                        (21) Symbolic evaluation [EQUIVALENT]
                        (22) Formula
                        (23) Hypothesis Lifting [EQUIVALENT]
                        (24) Formula
                        (25) Symbolic evaluation under hypothesis [SOUND]
                        (26) Formula
                        (27) Hypothesis Lifting [EQUIVALENT]
                        (28) Formula
                        (29) Hypothesis Lifting [EQUIVALENT]
                        (30) Formula
                        (31) Conditional Evaluation [EQUIVALENT]
                        (32) AND
                            (33) Formula
                                (34) Symbolic evaluation under hypothesis [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:
z2:sort[a34].(~(z2=nil)->del'(max(z2), z2)=true)

There are no hypotheses.




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

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm max(z2) generates the following cases:

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

There are no hypotheses.





2. Base Case:
Formula:
x11:sort[a33].(~(cons(x11, nil)=nil)->del'(max(cons(x11, nil)), cons(x11, nil))=true)

There are no hypotheses.





1. Step Case:
Formula:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].(~(cons(x25, cons(y16, xs10))=nil)->del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true)

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true





2. Step Case:
Formula:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].(~(cons(x25, cons(y16, xs10))=nil)->del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true)

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
(~(nil=nil)->del'(max(nil), nil)=true)

There are no hypotheses.




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

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

(5)
YES

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

(6)
Obligation:
Formula:
x11:sort[a33].(~(cons(x11, nil)=nil)->del'(max(cons(x11, nil)), cons(x11, nil))=true)

There are no hypotheses.




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

(7) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x11:sort[a33].del'(x11, cons(x11, nil))=true
----------------------------------------

(8)
Obligation:
Formula:
x11:sort[a33].del'(x11, cons(x11, nil))=true

There are no hypotheses.




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

(9) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a33] generates the following cases:



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

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a33].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true






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

(10)
Complex Obligation (AND)

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

(11)
Obligation:
Formula:
del'(0, cons(0, nil))=true

There are no hypotheses.




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

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

(13)
YES

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

(14)
Obligation:
Formula:
n:sort[a33].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true




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

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

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=true





Formula:
n:sort[a33].del'(s(n), nil)=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=false






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

(16)
Complex Obligation (AND)

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

(17)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=true




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

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

(19)
YES

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

(20)
Obligation:
Formula:
n:sort[a33].del'(s(n), nil)=true

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=false




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

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

(22)
Obligation:
Formula:
False

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=false




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

(23) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a33].((del'(n, cons(n, nil))=true/\equal_bool(eq(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=false




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

(24)
Obligation:
Formula:
n:sort[a33].((del'(n, cons(n, nil))=true/\equal_bool(eq(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=false




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

(25) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a33].~(equal_bool(eq(n, n), true)=false)

By using the following hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true

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

(26)
Obligation:
Formula:
n:sort[a33].~(equal_bool(eq(n, n), true)=false)

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true
n:sort[a33].equal_bool(eq(s(n), s(n)), true)=false




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

(27) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a33].(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false))

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true




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

(28)
Obligation:
Formula:
n:sort[a33].(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false))

Hypotheses:
n:sort[a33].del'(n, cons(n, nil))=true




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

(29) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a33].(del'(n, cons(n, nil))=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

There are no hypotheses.




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

(30)
Obligation:
Formula:
n:sort[a33].(del'(n, cons(n, nil))=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

There are no hypotheses.




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

(31) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a33].(true=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a33].equal_bool(eq(n, n), true)=true





Formula:
n:sort[a33].(del'(n, nil)=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a33].equal_bool(eq(n, n), true)=false






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

(32)
Complex Obligation (AND)

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

(33)
Obligation:
Formula:
n:sort[a33].(true=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a33].equal_bool(eq(n, n), true)=true




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

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

n:sort[a33].equal_bool(eq(n, n), true)=true

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

(35)
YES

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

(36)
Obligation:
Formula:
n:sort[a33].(del'(n, nil)=true->(equal_bool(eq(n, n), true)=false->~(equal_bool(eq(n, n), true)=false)))

Hypotheses:
n:sort[a33].equal_bool(eq(n, n), true)=false




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

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

(38)
YES

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

(39)
Obligation:
Formula:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].(~(cons(x25, cons(y16, xs10))=nil)->del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true)

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true




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

(40) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true
----------------------------------------

(41)
Obligation:
Formula:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true




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

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x25:sort[a33],xs10:sort[a34],y16:sort[a33].del'(max(cons(x25, xs10)), cons(x25, cons(y16, xs10)))=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true






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

(43)
Obligation:
Formula:
x25:sort[a33],xs10:sort[a34],y16:sort[a33].del'(max(cons(x25, xs10)), cons(x25, cons(y16, xs10)))=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true




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

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

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=true





Formula:
x25:sort[a33],xs10:sort[a34],y16:sort[a33].del'(max(cons(x25, xs10)), cons(y16, xs10))=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false






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

(45)
Complex Obligation (AND)

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

(46)
Obligation:
Formula:
true=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=true




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

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

(48)
YES

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

(49)
Obligation:
Formula:
x25:sort[a33],xs10:sort[a34],y16:sort[a33].del'(max(cons(x25, xs10)), cons(y16, xs10))=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false




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

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

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=true





Formula:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), xs10)=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=false






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

(51)
Complex Obligation (AND)

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

(52)
Obligation:
Formula:
true=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=true




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

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

(54)
YES

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

(55)
Obligation:
Formula:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), xs10)=true

Hypotheses:
x25:sort[a33],xs10:sort[a34].del'(max(cons(x25, xs10)), cons(x25, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=false




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

(56) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x25:sort[a33],xs10:sort[a34].(del'(max(cons(x25, xs10)), cons(x25, xs10))=true->del'(max(cons(x25, xs10)), xs10)=true)

Hypotheses:
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=false




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

(57)
Obligation:
Formula:
x25:sort[a33],xs10:sort[a34].(del'(max(cons(x25, xs10)), cons(x25, xs10))=true->del'(max(cons(x25, xs10)), xs10)=true)

Hypotheses:
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=false




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

(58) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
x25:sort[a33],xs10:sort[a34].(del'(max(cons(x25, xs10)), xs10)=true->del'(max(cons(x25, xs10)), xs10)=true)

Hypotheses:
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=false






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

(59)
Obligation:
Formula:
x25:sort[a33],xs10:sort[a34].(del'(max(cons(x25, xs10)), xs10)=true->del'(max(cons(x25, xs10)), xs10)=true)

Hypotheses:
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=true
x25:sort[a33],xs10:sort[a34].equal_bool(eq(max(cons(x25, xs10)), x25), true)=false
x25:sort[a33],xs10:sort[a34],y16:sort[a33].equal_bool(eq(max(cons(x25, xs10)), y16), true)=false




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

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

(61)
YES

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

(62)
Obligation:
Formula:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].(~(cons(x25, cons(y16, xs10))=nil)->del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true)

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false




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

(63) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true
----------------------------------------

(64)
Obligation:
Formula:
x25:sort[a33],y16:sort[a33],xs10:sort[a34].del'(max(cons(x25, cons(y16, xs10))), cons(x25, cons(y16, xs10)))=true

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false




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

(65) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y16:sort[a33],xs10:sort[a34],x25:sort[a33].del'(max(cons(y16, xs10)), cons(x25, cons(y16, xs10)))=true

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false






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

(66)
Obligation:
Formula:
y16:sort[a33],xs10:sort[a34],x25:sort[a33].del'(max(cons(y16, xs10)), cons(x25, cons(y16, xs10)))=true

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false




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

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

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false
y16:sort[a33],xs10:sort[a34],x25:sort[a33].equal_bool(eq(max(cons(y16, xs10)), x25), true)=true





Formula:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false
y16:sort[a33],xs10:sort[a34],x25:sort[a33].equal_bool(eq(max(cons(y16, xs10)), x25), true)=false






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

(68)
Complex Obligation (AND)

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

(69)
Obligation:
Formula:
true=true

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false
y16:sort[a33],xs10:sort[a34],x25:sort[a33].equal_bool(eq(max(cons(y16, xs10)), x25), true)=true




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

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

(71)
YES

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

(72)
Obligation:
Formula:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true

Hypotheses:
y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true
x25:sort[a33],y16:sort[a33].equal_bool(ge(x25, y16), true)=false
y16:sort[a33],xs10:sort[a34],x25:sort[a33].equal_bool(eq(max(cons(y16, xs10)), x25), true)=false




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

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

y16:sort[a33],xs10:sort[a34].del'(max(cons(y16, xs10)), cons(y16, xs10))=true

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

(74)
YES

(53) Complex Obligation (AND)

(54) Obligation:

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

SORT(cons(x0, x1)) → IF3(false, cons(x0, x1))

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

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

(55) DependencyGraphProof (EQUIVALENT transformation)

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

(56) TRUE

(57) Obligation:

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

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

Q is empty.

(58) Overlay + Local Confluence (EQUIVALENT transformation)

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

(59) Obligation:

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

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(60) DependencyPairsProof (EQUIVALENT transformation)

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

(61) Obligation:

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

DEL'(x81, cons(y55, xs37)) → IF2'(eq(x81, y55), x81, y55, xs37)
DEL'(x81, cons(y55, xs37)) → EQ(x81, y55)
IF2'(false, x163, y113, xs75) → DEL'(x163, xs75)
MAX(cons(x25, cons(y16, xs10))) → IF1(ge(x25, y16), x25, y16, xs10)
MAX(cons(x25, cons(y16, xs10))) → GE(x25, y16)
IF1(true, x39, y26, xs17) → MAX(cons(x39, xs17))
IF1(false, x53, y36, xs24) → MAX(cons(y36, xs24))
DEL(x81, cons(y55, xs37)) → IF2(eq(x81, y55), x81, y55, xs37)
DEL(x81, cons(y55, xs37)) → EQ(x81, y55)
EQ(s(x135), s(y93)) → EQ(x135, y93)
IF2(false, x163, y113, xs75) → DEL(x163, xs75)
GE(s(x205), s(y141)) → GE(x205, y141)
EQUAL_SORT[A33](s(x0), s(x1)) → EQUAL_SORT[A33](x0, x1)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → AND(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x0, x2)
EQUAL_SORT[A34](cons(x0, x1), cons(x2, x3)) → EQUAL_SORT[A34](x1, x3)

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(62) DependencyGraphProof (EQUIVALENT transformation)

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

(63) Complex Obligation (AND)

(64) Obligation:

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

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

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(65) 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.

(66) Obligation:

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

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

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

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(67) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(68) Obligation:

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

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

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

(69) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(70) YES

(71) Obligation:

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

EQUAL_SORT[A33](s(x0), s(x1)) → EQUAL_SORT[A33](x0, x1)

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(72) 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.

(73) Obligation:

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

EQUAL_SORT[A33](s(x0), s(x1)) → EQUAL_SORT[A33](x0, x1)

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

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(74) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(75) Obligation:

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

EQUAL_SORT[A33](s(x0), s(x1)) → EQUAL_SORT[A33](x0, x1)

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

(76) 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[A33](s(x0), s(x1)) → EQUAL_SORT[A33](x0, x1)
The graph contains the following edges 1 > 1, 2 > 2

(77) YES

(78) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(79) 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.

(80) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

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

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(81) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(82) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

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

(83) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

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

(84) YES

(85) Obligation:

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

EQ(s(x135), s(y93)) → EQ(x135, y93)

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(86) 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.

(87) Obligation:

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

EQ(s(x135), s(y93)) → EQ(x135, y93)

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

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(88) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(89) Obligation:

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

EQ(s(x135), s(y93)) → EQ(x135, y93)

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

(90) 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(x135), s(y93)) → EQ(x135, y93)
The graph contains the following edges 1 > 1, 2 > 2

(91) YES

(92) Obligation:

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

IF2(false, x163, y113, xs75) → DEL(x163, xs75)
DEL(x81, cons(y55, xs37)) → IF2(eq(x81, y55), x81, y55, xs37)

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(93) 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.

(94) Obligation:

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

IF2(false, x163, y113, xs75) → DEL(x163, xs75)
DEL(x81, cons(y55, xs37)) → IF2(eq(x81, y55), x81, y55, xs37)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(95) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(96) Obligation:

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

IF2(false, x163, y113, xs75) → DEL(x163, xs75)
DEL(x81, cons(y55, xs37)) → IF2(eq(x81, y55), x81, y55, xs37)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, 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.

(97) 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(x81, cons(y55, xs37)) → IF2(eq(x81, y55), x81, y55, xs37)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2(false, x163, y113, xs75) → DEL(x163, xs75)
The graph contains the following edges 2 >= 1, 4 >= 2

(98) YES

(99) Obligation:

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

IF1(true, x39, y26, xs17) → MAX(cons(x39, xs17))
MAX(cons(x25, cons(y16, xs10))) → IF1(ge(x25, y16), x25, y16, xs10)
IF1(false, x53, y36, xs24) → MAX(cons(y36, xs24))

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(100) 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.

(101) Obligation:

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

IF1(true, x39, y26, xs17) → MAX(cons(x39, xs17))
MAX(cons(x25, cons(y16, xs10))) → IF1(ge(x25, y16), x25, y16, xs10)
IF1(false, x53, y36, xs24) → MAX(cons(y36, xs24))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(102) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(103) Obligation:

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

IF1(true, x39, y26, xs17) → MAX(cons(x39, xs17))
MAX(cons(x25, cons(y16, xs10))) → IF1(ge(x25, y16), x25, y16, xs10)
IF1(false, x53, y36, xs24) → MAX(cons(y36, xs24))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(104) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF1(true, x39, y26, xs17) → MAX(cons(x39, xs17))
IF1(false, x53, y36, xs24) → MAX(cons(y36, xs24))

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 +
[ 0, 1 ]

· x₄

POL(true) =
/ 0 \

\ 1 /

POL(MAX(x₁)) = 0 +
[ 1, 0 ]

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁ +
/ 0 1 \

\ 0 1 /

· x₂

POL(ge(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 1 0 /

· x₁ +
/ 0 1 \

\ 1 0 /

· x₂

POL(false) =
/ 0 \

\ 1 /

POL(0) =
/ 0 \

\ 0 /

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

The following usable rules [FROCOS05] were oriented: none

(105) Obligation:

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

MAX(cons(x25, cons(y16, xs10))) → IF1(ge(x25, y16), x25, y16, xs10)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(106) DependencyGraphProof (EQUIVALENT transformation)

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

(107) TRUE

(108) Obligation:

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

IF2'(false, x163, y113, xs75) → DEL'(x163, xs75)
DEL'(x81, cons(y55, xs37)) → IF2'(eq(x81, y55), x81, y55, xs37)

The TRS R consists of the following rules:

del'(x67, nil) → false
del'(x81, cons(y55, xs37)) → if2'(eq(x81, y55), x81, y55, xs37)
if2'(true, x149, y103, xs68) → true
if2'(false, x163, y113, xs75) → del'(x163, xs75)
max(nil) → 0
max(cons(x11, nil)) → x11
max(cons(x25, cons(y16, xs10))) → if1(ge(x25, y16), x25, y16, xs10)
if1(true, x39, y26, xs17) → max(cons(x39, xs17))
if1(false, x53, y36, xs24) → max(cons(y36, xs24))
del(x67, nil) → nil
del(x81, cons(y55, xs37)) → if2(eq(x81, y55), x81, y55, xs37)
eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)
if2(true, x149, y103, xs68) → xs68
if2(false, x163, y113, xs75) → cons(y113, del(x163, xs75))
ge(x177, 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
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[a33](0, 0) → true
equal_sort[a33](0, s(x0)) → false
equal_sort[a33](s(x0), 0) → false
equal_sort[a33](s(x0), s(x1)) → equal_sort[a33](x0, x1)
equal_sort[a34](nil, nil) → true
equal_sort[a34](nil, cons(x0, x1)) → false
equal_sort[a34](cons(x0, x1), nil) → false
equal_sort[a34](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a34](x0, x2), equal_sort[a34](x1, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(109) 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.

(110) Obligation:

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

IF2'(false, x163, y113, xs75) → DEL'(x163, xs75)
DEL'(x81, cons(y55, xs37)) → IF2'(eq(x81, y55), x81, y55, xs37)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, y93)

The set Q consists of the following terms:

del'(x0, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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))
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

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

(111) 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, nil)
del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
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[a33](0, 0)
equal_sort[a33](0, s(x0))
equal_sort[a33](s(x0), 0)
equal_sort[a33](s(x0), s(x1))
equal_sort[a34](nil, nil)
equal_sort[a34](nil, cons(x0, x1))
equal_sort[a34](cons(x0, x1), nil)
equal_sort[a34](cons(x0, x1), cons(x2, x3))
equal_sort[a63](witness_sort[a63], witness_sort[a63])

(112) Obligation:

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

IF2'(false, x163, y113, xs75) → DEL'(x163, xs75)
DEL'(x81, cons(y55, xs37)) → IF2'(eq(x81, y55), x81, y55, xs37)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(y74)) → false
eq(s(x121), 0) → false
eq(s(x135), s(y93)) → eq(x135, 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.

(113) 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'(x81, cons(y55, xs37)) → IF2'(eq(x81, y55), x81, y55, xs37)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF2'(false, x163, y113, xs75) → DEL'(x163, xs75)
The graph contains the following edges 2 >= 1, 4 >= 2