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 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 QDPSizeChangeProof (⇔)
    ↳27 TRUE
  ↳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 Induction-Processor (⇐)
    ↳43 AND
      ↳44 QDP
        ↳45 PisEmptyProof (⇔)
        ↳46 TRUE
      ↳47 QTRS
        ↳48 Overlay + Local Confluence (⇔)
        ↳49 QTRS
        ↳50 DependencyPairsProof (⇔)
        ↳51 QDP
        ↳52 DependencyGraphProof (⇔)
        ↳53 AND
          ↳54 QDP
            ↳55 UsableRulesProof (⇔)
            ↳56 QDP
            ↳57 QReductionProof (⇔)
            ↳58 QDP
            ↳59 QDPSizeChangeProof (⇔)
            ↳60 TRUE
          ↳61 QDP
            ↳62 UsableRulesProof (⇔)
            ↳63 QDP
            ↳64 QReductionProof (⇔)
            ↳65 QDP
            ↳66 QDPSizeChangeProof (⇔)
            ↳67 TRUE
          ↳68 QDP
            ↳69 UsableRulesProof (⇔)
            ↳70 QDP
            ↳71 QReductionProof (⇔)
            ↳72 QDP
            ↳73 QDPSizeChangeProof (⇔)
            ↳74 TRUE
          ↳75 QDP
            ↳76 UsableRulesProof (⇔)
            ↳77 QDP
            ↳78 QReductionProof (⇔)
            ↳79 QDP
            ↳80 QDPSizeChangeProof (⇔)
            ↳81 TRUE
          ↳82 QDP
            ↳83 UsableRulesProof (⇔)
            ↳84 QDP
            ↳85 QReductionProof (⇔)
            ↳86 QDP
            ↳87 QDPSizeChangeProof (⇔)
            ↳88 TRUE
          ↳89 QDP
            ↳90 UsableRulesProof (⇔)
            ↳91 QDP
            ↳92 QReductionProof (⇔)
            ↳93 QDP
            ↳94 QDPOrderProof (⇔)
            ↳95 QDP
            ↳96 DependencyGraphProof (⇔)
            ↳97 TRUE
          ↳98 QDP
            ↳99 UsableRulesProof (⇔)
            ↳100 QDP
            ↳101 QReductionProof (⇔)
            ↳102 QDP
            ↳103 QDPSizeChangeProof (⇔)
            ↳104 TRUE

(0) Obligation:

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

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

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

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

The set Q consists of the following terms:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

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

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

(10) QReductionProof (EQUIVALENT transformation)

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

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

(11) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(13) TRUE

(14) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

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

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

(17) QReductionProof (EQUIVALENT transformation)

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

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

(18) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(20) TRUE

(21) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(22) UsableRulesProof (EQUIVALENT transformation)

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

(23) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(24) QReductionProof (EQUIVALENT transformation)

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

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

(25) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

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

(27) TRUE

(28) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(29) UsableRulesProof (EQUIVALENT transformation)

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

(30) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(31) QReductionProof (EQUIVALENT transformation)

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

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

(32) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MAX(cons(x, cons(y, xs))) → IF1(ge(x, y), x, y, xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(38) UsableRulesProof (EQUIVALENT transformation)

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

(39) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(40) QReductionProof (EQUIVALENT transformation)

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

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

(41) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(42) Induction-Processor (SOUND transformation)


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


This order was computed:
Polynomial interpretation [POLO]:

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

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


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


The transformed set:
del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

(43) Complex Obligation (AND)

(44) Obligation:

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

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

The set Q consists of the following terms:

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

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

Q is empty.

(48) Overlay + Local Confluence (EQUIVALENT transformation)

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

(49) Obligation:

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

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(50) DependencyPairsProof (EQUIVALENT transformation)

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

(51) Obligation:

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

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

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(52) DependencyGraphProof (EQUIVALENT transformation)

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

(53) Complex Obligation (AND)

(54) Obligation:

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

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

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(55) UsableRulesProof (EQUIVALENT transformation)

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

(56) Obligation:

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

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

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(57) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(58) Obligation:

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

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

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

(59) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

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

(60) TRUE

(61) Obligation:

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

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

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(62) UsableRulesProof (EQUIVALENT transformation)

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

(63) Obligation:

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

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

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(64) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(65) Obligation:

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

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

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

(66) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(67) TRUE

(68) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(69) UsableRulesProof (EQUIVALENT transformation)

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

(70) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(71) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(72) Obligation:

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

GE(s(x205), s(y141)) → GE(x205, y141)

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

(73) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(74) TRUE

(75) Obligation:

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

EQ(s(x108), s(y75)) → EQ(x108, y75)

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(76) UsableRulesProof (EQUIVALENT transformation)

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

(77) Obligation:

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

EQ(s(x108), s(y75)) → EQ(x108, y75)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(78) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(79) Obligation:

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

EQ(s(x108), s(y75)) → EQ(x108, y75)

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

(80) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

(81) TRUE

(82) Obligation:

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

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

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(83) UsableRulesProof (EQUIVALENT transformation)

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

(84) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(85) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(86) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(87) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

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

(88) TRUE

(89) Obligation:

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

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

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(90) UsableRulesProof (EQUIVALENT transformation)

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

(91) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(92) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(93) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(94) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MAX(cons(x12, cons(y7, xs4))) → IF1(ge(x12, y7), x12, y7, xs4)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(95) Obligation:

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

IF1(true, x26, y17, xs11) → MAX(cons(x26, xs11))
IF1(false, x40, y27, xs18) → MAX(cons(y27, xs18))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(96) DependencyGraphProof (EQUIVALENT transformation)

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

(97) TRUE

(98) Obligation:

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

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

The TRS R consists of the following rules:

del'(x54, cons(y37, xs25)) → if2'(eq(x54, y37), x54, y37, xs25)
if2'(true, x122, y85, xs56) → true
if2'(false, x136, y95, xs63) → del'(x136, xs63)
del'(x150, nil) → false
max(cons(x, nil)) → x
max(cons(x12, cons(y7, xs4))) → if1(ge(x12, y7), x12, y7, xs4)
if1(true, x26, y17, xs11) → max(cons(x26, xs11))
if1(false, x40, y27, xs18) → max(cons(y27, xs18))
del(x54, cons(y37, xs25)) → if2(eq(x54, y37), x54, y37, xs25)
eq(0, 0) → true
eq(0, s(y56)) → false
eq(s(x94), 0) → false
eq(s(x108), s(y75)) → eq(x108, y75)
if2(true, x122, y85, xs56) → xs56
if2(false, x136, y95, xs63) → cons(y95, del(x136, xs63))
del(x150, nil) → nil
ge(0, 0) → true
ge(s(x177), 0) → true
ge(0, s(x191)) → false
ge(s(x205), s(y141)) → ge(x205, y141)
max(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60]) → true

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(99) UsableRulesProof (EQUIVALENT transformation)

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

(100) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

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

(101) QReductionProof (EQUIVALENT transformation)

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

del'(x0, cons(x1, x2))
if2'(true, x0, x1, x2)
if2'(false, x0, x1, x2)
del'(x0, nil)
max(cons(x0, nil))
max(cons(x0, cons(x1, x2)))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
del(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
del(x0, nil)
ge(0, 0)
ge(s(x0), 0)
ge(0, s(x0))
ge(s(x0), s(x1))
max(nil)
equal_bool(true, false)
equal_bool(false, true)
equal_bool(true, true)
equal_bool(false, false)
and(true, x0)
and(false, x0)
or(true, x0)
or(false, x0)
not(false)
not(true)
isa_true(true)
isa_true(false)
isa_false(true)
isa_false(false)
equal_sort[a0](0, 0)
equal_sort[a0](0, s(x0))
equal_sort[a0](s(x0), 0)
equal_sort[a0](s(x0), s(x1))
equal_sort[a34](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[a60](witness_sort[a60], witness_sort[a60])

(102) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(103) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

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

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

(104) TRUE