Termination w.r.t. Q of the following Term Rewriting System could be proven:

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SORTITER(x, y) → EMPTY(x)
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
IF(false, x, y, z) → TAIL(x)
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF(false, x, y, z) → REPLACE(min(x), head(x), tail(x))
LE(s(n), s(m)) → LE(n, m)
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)
REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
EQ(s(n), s(m)) → EQ(n, m)
MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
SORT(x) → SORTITER(x, nil)
SORTITER(x, y) → MIN(x)
IF(false, x, y, z) → HEAD(x)
MIN(cons(n, cons(m, x))) → LE(n, m)
REPLACE(n, m, cons(k, x)) → EQ(n, k)
IF(false, x, y, z) → MIN(x)
SORTITER(x, y) → IF(empty(x), x, y, append(y, cons(min(x), nil)))

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

SORTITER(x, y) → EMPTY(x)
IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
IF(false, x, y, z) → TAIL(x)
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
IF(false, x, y, z) → REPLACE(min(x), head(x), tail(x))
LE(s(n), s(m)) → LE(n, m)
IF_REPLACE(false, n, m, cons(k, x)) → REPLACE(n, m, x)
REPLACE(n, m, cons(k, x)) → IF_REPLACE(eq(n, k), n, m, cons(k, x))
IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
EQ(s(n), s(m)) → EQ(n, m)
MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
SORT(x) → SORTITER(x, nil)
SORTITER(x, y) → MIN(x)
IF(false, x, y, z) → HEAD(x)
MIN(cons(n, cons(m, x))) → LE(n, m)
REPLACE(n, m, cons(k, x)) → EQ(n, k)
IF(false, x, y, z) → MIN(x)
SORTITER(x, y) → IF(empty(x), x, y, append(y, cons(min(x), nil)))

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 9 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, 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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

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

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
min(cons(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MIN(cons(n, cons(m, x))) → IF_MIN(le(n, m), cons(n, cons(m, x)))
The remaining pairs can at least be oriented weakly.

IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, x))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(IF_MIN(x1, x2)) = 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 [17] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ DependencyGraphProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

IF_MIN(false, cons(n, cons(m, x))) → MIN(cons(m, x))
IF_MIN(true, cons(n, cons(m, x))) → MIN(cons(n, 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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ QDP

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(x, y) → IF(empty(x), x, y, append(y, cons(min(x), nil)))

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(x, nil)) → 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))
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))
empty(nil) → true
empty(cons(n, x)) → false
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → x
sort(x) → sortIter(x, nil)
sortIter(x, y) → if(empty(x), x, y, append(y, cons(min(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → sortIter(replace(min(x), head(x), tail(x)), z)

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(x, y) → IF(empty(x), x, y, append(y, cons(min(x), nil)))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
min(cons(x, nil)) → x
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))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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))

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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

sort(x0)
sortIter(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ Narrowing

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(x, y) → IF(empty(x), x, y, append(y, cons(min(x), nil)))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
min(cons(x, nil)) → x
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))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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))

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule SORTITER(x, y) → IF(empty(x), x, y, append(y, cons(min(x), nil))) at position [0] we obtained the following new rules:

SORTITER(nil, y1) → IF(true, nil, y1, append(y1, cons(min(nil), nil)))
SORTITER(cons(x0, x1), y1) → IF(false, cons(x0, x1), y1, append(y1, cons(min(cons(x0, x1)), nil)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ DependencyGraphProof

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

SORTITER(nil, y1) → IF(true, nil, y1, append(y1, cons(min(nil), nil)))
IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(cons(x0, x1), y1) → IF(false, cons(x0, x1), y1, append(y1, cons(min(cons(x0, x1)), nil)))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
min(cons(x, nil)) → x
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))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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))

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
                                ↳ UsableRulesProof

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(cons(x0, x1), y1) → IF(false, cons(x0, x1), y1, append(y1, cons(min(cons(x0, x1)), nil)))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
min(cons(x, nil)) → x
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))
le(0, m) → true
le(s(n), 0) → false
le(s(n), s(m)) → le(n, m)
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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))

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(cons(x0, x1), y1) → IF(false, cons(x0, x1), y1, append(y1, cons(min(cons(x0, x1)), nil)))

The TRS R consists of the following rules:

min(cons(x, nil)) → x
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))
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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(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))
empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
QDP
                                        ↳ QDPOrderProof

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
SORTITER(cons(x0, x1), y1) → IF(false, cons(x0, x1), y1, append(y1, cons(min(cons(x0, x1)), nil)))

The TRS R consists of the following rules:

min(cons(x, nil)) → x
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))
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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(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))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


SORTITER(cons(x0, x1), y1) → IF(false, cons(x0, x1), y1, append(y1, cons(min(cons(x0, x1)), nil)))
The remaining pairs can at least be oriented weakly.

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( tail(x1) ) =
/0\
\0/
+
/01\
\10/
·x1

M( eq(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( le(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( head(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( true ) =
/0\
\0/

M( append(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\01/
·x2

M( 0 ) =
/0\
\0/

M( cons(x1, x2) ) =
/0\
\1/
+
/00\
\00/
·x1+
/01\
\11/
·x2

M( false ) =
/0\
\0/

M( s(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( if_replace(x1, ..., x4) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2+
/00\
\00/
·x3+
/10\
\01/
·x4

M( min(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( nil ) =
/0\
\0/

M( if_min(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( replace(x1, ..., x3) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2+
/10\
\01/
·x3

Tuple symbols:
M( SORTITER(x1, x2) ) = 0+
[0,1]
·x1+
[0,0]
·x2

M( IF(x1, ..., x4) ) = 0+
[0,0]
·x1+
[1,0]
·x2+
[0,0]
·x3+
[0,0]
·x4


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
                                      ↳ QDP
                                        ↳ QDPOrderProof
QDP
                                            ↳ DependencyGraphProof

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

IF(false, x, y, z) → SORTITER(replace(min(x), head(x), tail(x)), z)

The TRS R consists of the following rules:

min(cons(x, nil)) → x
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))
head(cons(n, x)) → n
tail(nil) → nil
tail(cons(n, x)) → 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(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))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.