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 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 DependencyGraphProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 UsableRulesProof (⇔)
    ↳25 QDP
    ↳26 QReductionProof (⇔)
    ↳27 QDP
    ↳28 QDPSizeChangeProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 UsableRulesProof (⇔)
    ↳32 QDP
    ↳33 QReductionProof (⇔)
    ↳34 QDP
    ↳35 QDPSizeChangeProof (⇔)
    ↳36 TRUE
  ↳37 QDP
    ↳38 UsableRulesProof (⇔)
    ↳39 QDP
    ↳40 QReductionProof (⇔)
    ↳41 QDP
    ↳42 QDPOrderProof (⇔)
    ↳43 QDP
    ↳44 PisEmptyProof (⇔)
    ↳45 TRUE

(0) Obligation:

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

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

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:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

EQ(s(n), s(m)) → EQ(n, m)
LE(s(n), s(m)) → LE(n, m)
MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
MIN(cons(n, cons(m, x))) → LE(n, m)
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
REPLACE(n, m, cons(k, x)) → EQ(n, k)
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)
SORT(cons(n, x)) → MIN(cons(n, x))
SORT(cons(n, x)) → SORT(replace(min(cons(n, x)), n, x))
SORT(cons(n, x)) → REPLACE(min(cons(n, x)), n, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(n), s(m)) → LE(n, m)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

LE(s(n), s(m)) → LE(n, m)

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

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(10) QReductionProof (EQUIVALENT transformation)

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

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

(11) Obligation:

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

LE(s(n), s(m)) → LE(n, m)

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:

  • LE(s(n), s(m)) → LE(n, m)
    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:

MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))

The TRS R consists of the following rules:

le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(17) QReductionProof (EQUIVALENT transformation)

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

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

(18) Obligation:

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

MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))

The TRS R consists of the following rules:

le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF_MIN(x1, x2)) = 1 + x2   
POL(MIN(x1)) = 1 + x1   
POL(cons(x1, x2)) = 1 + x2   
POL(false) = 0   
POL(le(x1, x2)) = 0   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))

The TRS R consists of the following rules:

le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

EQ(s(n), s(m)) → EQ(n, m)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(24) 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.

(25) Obligation:

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

EQ(s(n), s(m)) → EQ(n, m)

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

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(26) 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].

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

(27) Obligation:

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

EQ(s(n), s(m)) → EQ(n, m)

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

(28) 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(n), s(m)) → EQ(n, m)
    The graph contains the following edges 1 > 1, 2 > 2

(29) TRUE

(30) Obligation:

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

REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(31) 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.

(32) Obligation:

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

REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

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

(33) 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, x1))

(34) Obligation:

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

REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)

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.

(35) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)
    The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3

  • REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
    The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4

(36) TRUE

(37) Obligation:

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

SORT(cons(n, x)) → SORT(replace(min(cons(n, x)), n, x))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
sort(nil) → nil
sort(cons(n, x)) → cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, 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(n, x)) → SORT(replace(min(cons(n, x)), n, x))

The TRS R consists of the following rules:

min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))
sort(nil)
sort(cons(x0, 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(n, x)) → SORT(replace(min(cons(n, x)), n, x))

The TRS R consists of the following rules:

min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SORT(cons(n, x)) → SORT(replace(min(cons(n, x)), n, x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(SORT(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(eq(x1, x2)) = 0   
POL(false) = 0   
POL(if_min(x1, x2)) = 0   
POL(if_replace(x1, x2, x3, x4)) = x4   
POL(le(x1, x2)) = 0   
POL(min(x1)) = 0   
POL(nil) = 0   
POL(replace(x1, x2, x3)) = x3   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
if_replace(true, n, m, cons(k, x)) → cons(m, x)

(43) Obligation:

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

min(cons(0, nil)) → 0
min(cons(s(n), nil)) → s(n)
if_min(false, cons(n, cons(m, x))) → min(cons(m, x))
min(cons(n, cons(m, x))) → if_min(le(n, m), cons(n, cons(m, x)))
if_min(true, cons(n, cons(m, x))) → min(cons(n, x))
replace(n, m, nil) → nil
replace(n, m, cons(k, x)) → if_replace(eq(n, k), n, m, cons(k, x))
eq(0, 0) → true
eq(0, s(m)) → false
eq(s(n), 0) → false
eq(s(n), s(m)) → eq(n, m)
if_replace(true, n, m, cons(k, x)) → cons(m, x)
if_replace(false, n, m, cons(k, x)) → cons(k, replace(n, m, x))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
if_min(true, cons(x0, cons(x1, x2)))
if_min(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
if_replace(true, x0, x1, cons(x2, x3))
if_replace(false, x0, x1, cons(x2, x3))

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

(44) PisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE