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 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 DependencyGraphProof (⇔)
    ↳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 QDPSizeChangeProof (⇔)
    ↳43 TRUE
  ↳44 QDP
    ↳45 UsableRulesProof (⇔)
    ↳46 QDP
    ↳47 QReductionProof (⇔)
    ↳48 QDP
      ↳49 Instantiation (⇔)
        ↳50 QDP
      ↳51 Narrowing (⇔)
        ↳52 QDP
        ↳53 DependencyGraphProof (⇔)
        ↳54 QDP
        ↳55 UsableRulesProof (⇔)
        ↳56 QDP
        ↳57 QReductionProof (⇔)
        ↳58 QDP
        ↳59 Narrowing (⇔)
        ↳60 QDP
        ↳61 Rewriting (⇔)
        ↳62 QDP
        ↳63 Rewriting (⇔)
        ↳64 QDP
        ↳65 Rewriting (⇔)
        ↳66 QDP
        ↳67 Rewriting (⇔)
        ↳68 QDP
        ↳69 Rewriting (⇔)
        ↳70 QDP
        ↳71 Narrowing (⇔)
        ↳72 QDP
        ↳73 DependencyGraphProof (⇔)
        ↳74 QDP
        ↳75 UsableRulesProof (⇔)
        ↳76 QDP
        ↳77 QReductionProof (⇔)
        ↳78 QDP
        ↳79 Rewriting (⇔)
        ↳80 QDP
        ↳81 Instantiation (⇔)
        ↳82 QDP
        ↳83 DependencyGraphProof (⇔)
        ↳84 QDP
        ↳85 Rewriting (⇔)
        ↳86 QDP
        ↳87 UsableRulesProof (⇔)
        ↳88 QDP
        ↳89 QReductionProof (⇔)
        ↳90 QDP
        ↳91 QDPOrderProof (⇔)
        ↳92 QDP
        ↳93 UsableRulesProof (⇔)
        ↳94 QDP
        ↳95 QReductionProof (⇔)
        ↳96 QDP
        ↳97 QDPSizeChangeProof (⇔)
        ↳98 TRUE

(0) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), z)

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(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(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(x), s(y)) → EQ(x, y)
LE(s(x), s(y)) → LE(x, y)
APP(add(n, x), y) → APP(x, y)
MIN(add(n, add(m, x))) → IF_MIN(le(n, m), add(n, add(m, x)))
MIN(add(n, add(m, x))) → LE(n, m)
IF_MIN(true, add(n, add(m, x))) → MIN(add(n, x))
IF_MIN(false, add(n, add(m, x))) → MIN(add(m, x))
RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
RM(n, add(m, x)) → EQ(n, m)
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)
MINSORT(x) → MINS(x, nil, nil)
MINS(x, y, z) → IF(null(x), x, y, z)
MINS(x, y, z) → NULL(x)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF(false, x, y, z) → EQ(head(x), min(x))
IF(false, x, y, z) → HEAD(x)
IF(false, x, y, z) → MIN(x)
IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
IF2(true, x, y, z) → APP(rm(head(x), tail(x)), y)
IF2(true, x, y, z) → RM(head(x), tail(x))
IF2(true, x, y, z) → HEAD(x)
IF2(true, x, y, z) → TAIL(x)
IF2(true, x, y, z) → APP(z, add(head(x), nil))
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(false, x, y, z) → TAIL(x)
IF2(false, x, y, z) → HEAD(x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 14 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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:

APP(add(n, x), y) → APP(x, y)

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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(11) Obligation:

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

APP(add(n, x), y) → APP(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:

  • APP(add(n, x), y) → APP(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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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:

LE(s(x), s(y)) → LE(x, y)

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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(18) Obligation:

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

LE(s(x), s(y)) → LE(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:

  • LE(s(x), s(y)) → LE(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:

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

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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:

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

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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].

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(25) Obligation:

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

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

The TRS R consists of the following rules:

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

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.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(MIN(x1)) =
/0\
\1/
 +
/11\
\11/
·x1

POL(add(x1, x2)) =
/1\
\0/
 +
/00\
\00/
·x1 +
/01\
\10/
·x2

POL(IF_MIN(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/11\
\11/
·x2

POL(le(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/10\
\10/
·x2

POL(true) =
/1\
\1/

POL(false) =
/1\
\0/

POL(s(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

POL(0) =
/1\
\1/

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

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

The TRS R consists of the following rules:

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

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.

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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:

EQ(s(x), s(y)) → EQ(x, y)

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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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].

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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(34) 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.

(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:

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

(36) TRUE

(37) Obligation:

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

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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:

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → 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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(41) Obligation:

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

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → 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.

(42) 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:

  • RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
    The graph contains the following edges 1 >= 2, 2 >= 3

  • IF_RM(true, n, add(m, x)) → RM(n, x)
    The graph contains the following edges 2 >= 1, 3 > 2

  • IF_RM(false, n, add(m, x)) → RM(n, x)
    The graph contains the following edges 2 >= 1, 3 > 2

(43) TRUE

(44) Obligation:

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

IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
MINS(x, y, z) → IF(null(x), x, y, z)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(x) → mins(x, nil, nil)
mins(x, y, z) → if(null(x), x, y, z)
if(true, x, y, z) → z
if(false, x, y, z) → if2(eq(head(x), min(x)), x, y, z)
if2(true, x, y, z) → mins(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
if2(false, x, y, z) → mins(tail(x), add(head(x), y), 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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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

(45) UsableRulesProof (EQUIVALENT transformation)

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

(46) Obligation:

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

IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
MINS(x, y, z) → IF(null(x), x, y, z)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

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

(47) QReductionProof (EQUIVALENT transformation)

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

minsort(x0)
mins(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(48) Obligation:

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

IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
MINS(x, y, z) → IF(null(x), x, y, z)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(49) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule MINS(x, y, z) → IF(null(x), x, y, z) we obtained the following new rules [LPAR04]:

MINS(y_3, nil, y_5) → IF(null(y_3), y_3, nil, y_5)
MINS(y_0, add(y_1, z1), z2) → IF(null(y_0), y_0, add(y_1, z1), z2)

(50) Obligation:

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

IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
MINS(y_3, nil, y_5) → IF(null(y_3), y_3, nil, y_5)
MINS(y_0, add(y_1, z1), z2) → IF(null(y_0), y_0, add(y_1, z1), z2)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(51) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MINS(x, y, z) → IF(null(x), x, y, z) at position [0] we obtained the following new rules [LPAR04]:

MINS(nil, y1, y2) → IF(true, nil, y1, y2)
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)

(52) Obligation:

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

IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
MINS(nil, y1, y2) → IF(true, nil, y1, y2)
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(53) DependencyGraphProof (EQUIVALENT transformation)

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

(54) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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:

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
null(nil)
null(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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].

null(nil)
null(add(x0, x1))

(58) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil)))
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(59) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF2(true, x, y, z) → MINS(app(rm(head(x), tail(x)), y), nil, app(z, add(head(x), nil))) at position [0] we obtained the following new rules [LPAR04]:

IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, tail(add(x0, x1))), y1), nil, app(y2, add(head(add(x0, x1)), nil)))
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(head(add(x0, x1)), x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))
IF2(true, nil, y1, y2) → MINS(app(rm(head(nil), nil), y1), nil, app(y2, add(head(nil), nil)))

(60) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, tail(add(x0, x1))), y1), nil, app(y2, add(head(add(x0, x1)), nil)))
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(head(add(x0, x1)), x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))
IF2(true, nil, y1, y2) → MINS(app(rm(head(nil), nil), y1), nil, app(y2, add(head(nil), nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(61) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, tail(add(x0, x1))), y1), nil, app(y2, add(head(add(x0, x1)), nil))) at position [0,0,1] we obtained the following new rules [LPAR04]:

IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))

(62) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(head(add(x0, x1)), x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))
IF2(true, nil, y1, y2) → MINS(app(rm(head(nil), nil), y1), nil, app(y2, add(head(nil), nil)))
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(63) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(head(add(x0, x1)), x1), y1), nil, app(y2, add(head(add(x0, x1)), nil))) at position [0,0,0] we obtained the following new rules [LPAR04]:

IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))

(64) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(true, nil, y1, y2) → MINS(app(rm(head(nil), nil), y1), nil, app(y2, add(head(nil), nil)))
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(65) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, nil, y1, y2) → MINS(app(rm(head(nil), nil), y1), nil, app(y2, add(head(nil), nil))) at position [0,0] we obtained the following new rules [LPAR04]:

IF2(true, nil, y1, y2) → MINS(app(nil, y1), nil, app(y2, add(head(nil), nil)))

(66) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), nil)))
IF2(true, nil, y1, y2) → MINS(app(nil, y1), nil, app(y2, add(head(nil), nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(67) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), nil))) at position [2,1,0] we obtained the following new rules [LPAR04]:

IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))

(68) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(true, nil, y1, y2) → MINS(app(nil, y1), nil, app(y2, add(head(nil), nil)))
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(69) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, nil, y1, y2) → MINS(app(nil, y1), nil, app(y2, add(head(nil), nil))) at position [0] we obtained the following new rules [LPAR04]:

IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))

(70) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(71) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF2(false, x, y, z) → MINS(tail(x), add(head(x), y), z) at position [0] we obtained the following new rules [LPAR04]:

IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(head(add(x0, x1)), y1), y2)
IF2(false, nil, y1, y2) → MINS(nil, add(head(nil), y1), y2)

(72) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(head(add(x0, x1)), y1), y2)
IF2(false, nil, y1, y2) → MINS(nil, add(head(nil), y1), y2)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(73) DependencyGraphProof (EQUIVALENT transformation)

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

(74) Obligation:

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

IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(head(add(x0, x1)), y1), y2)

The TRS R consists of the following rules:

tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(75) 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.

(76) Obligation:

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

IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(head(add(x0, x1)), y1), y2)

The TRS R consists of the following rules:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
tail(add(x0, x1))
tail(nil)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(77) 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].

tail(add(x0, x1))
tail(nil)

(78) Obligation:

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

IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(head(add(x0, x1)), y1), y2)

The TRS R consists of the following rules:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(79) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(head(add(x0, x1)), y1), y2) at position [1,0] we obtained the following new rules [LPAR04]:

IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)

(80) Obligation:

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

IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)

The TRS R consists of the following rules:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(81) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, x, y, z) → IF2(eq(head(x), min(x)), x, y, z) we obtained the following new rules [LPAR04]:

IF(false, add(z0, z1), z2, z3) → IF2(eq(head(add(z0, z1)), min(add(z0, z1))), add(z0, z1), z2, z3)

(82) Obligation:

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

IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(head(add(z0, z1)), min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(83) DependencyGraphProof (EQUIVALENT transformation)

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

(84) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(head(add(z0, z1)), min(add(z0, z1))), add(z0, z1), z2, z3)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)

The TRS R consists of the following rules:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(85) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(false, add(z0, z1), z2, z3) → IF2(eq(head(add(z0, z1)), min(add(z0, z1))), add(z0, z1), z2, z3) at position [0,0] we obtained the following new rules [LPAR04]:

IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

(86) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(87) 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.

(88) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
head(add(x0, x1))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(89) 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].

head(add(x0, x1))

(90) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(91) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, nil)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(IF(x1, x2, x3, x4)) = x2 + x3   
POL(IF2(x1, x2, x3, x4)) = x2 + x3   
POL(MINS(x1, x2, x3)) = x1 + x2   
POL(add(x1, x2)) = 1 + x2   
POL(app(x1, x2)) = x1 + x2   
POL(eq(x1, x2)) = 0   
POL(false) = 0   
POL(if_min(x1, x2)) = 0   
POL(if_rm(x1, x2, x3)) = x3   
POL(le(x1, x2)) = 1 + x2   
POL(min(x1)) = 0   
POL(nil) = 0   
POL(rm(x1, x2)) = x2   
POL(s(x1)) = x1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
rm(n, nil) → nil
app(add(n, x), y) → add(n, app(x, y))
app(nil, y) → y
if_rm(false, n, add(m, x)) → add(m, rm(n, x))

(92) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
if_rm(false, n, add(m, x)) → add(m, rm(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))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(93) UsableRulesProof (EQUIVALENT transformation)

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

(94) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

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

(95) QReductionProof (EQUIVALENT transformation)

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

app(nil, x0)
app(add(x0, x1), x2)
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))

(96) Obligation:

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

MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

The TRS R consists of the following rules:

min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(false, add(n, add(m, x))) → min(add(m, x))
if_min(true, add(n, add(m, x))) → min(add(n, x))
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))

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

(97) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)
    The graph contains the following edges 2 >= 2, 3 >= 3, 4 >= 4

  • IF2(false, add(x0, x1), y1, y2) → MINS(x1, add(x0, y1), y2)
    The graph contains the following edges 2 > 1, 4 >= 3

  • MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)
    The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4

(98) TRUE