YES Termination w.r.t. Q proof of AProVE_07_thiemann02.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 7 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPOrderProof (⇔, 14 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QReductionProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QDPSizeChangeProof (⇔, 0 ms)
        ↳43 YES
    ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QReductionProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 TransformationProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 DependencyGraphProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 UsableRulesProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 QReductionProof (⇔, 0 ms)
        ↳56 QDP
        ↳57 TransformationProof (⇔, 0 ms)
        ↳58 QDP
        ↳59 TransformationProof (⇔, 0 ms)
        ↳60 QDP
        ↳61 TransformationProof (⇔, 0 ms)
        ↳62 QDP
        ↳63 TransformationProof (⇔, 0 ms)
        ↳64 QDP
        ↳65 TransformationProof (⇔, 0 ms)
        ↳66 QDP
        ↳67 TransformationProof (⇔, 0 ms)
        ↳68 QDP
        ↳69 TransformationProof (⇔, 0 ms)
        ↳70 QDP
        ↳71 DependencyGraphProof (⇔, 0 ms)
        ↳72 QDP
        ↳73 UsableRulesProof (⇔, 0 ms)
        ↳74 QDP
        ↳75 QReductionProof (⇔, 0 ms)
        ↳76 QDP
        ↳77 TransformationProof (⇔, 0 ms)
        ↳78 QDP
        ↳79 TransformationProof (⇔, 0 ms)
        ↳80 QDP
        ↳81 DependencyGraphProof (⇔, 0 ms)
        ↳82 QDP
        ↳83 TransformationProof (⇔, 0 ms)
        ↳84 QDP
        ↳85 UsableRulesProof (⇔, 0 ms)
        ↳86 QDP
        ↳87 QReductionProof (⇔, 0 ms)
        ↳88 QDP
        ↳89 QDPOrderProof (⇔, 53 ms)
        ↳90 QDP
        ↳91 UsableRulesProof (⇔, 0 ms)
        ↳92 QDP
        ↳93 QReductionProof (⇔, 0 ms)
        ↳94 QDP
        ↳95 QDPSizeChangeProof (⇔, 0 ms)
        ↳96 YES

(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) YES

(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) YES

(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,JAR06].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( IF_MIN(x1, x2) ) = max{0, 2x1 + x2 - 2}

POL( le(x1, x2) ) = 2

POL( 0 ) = 0

POL( true ) = 2

POL( s(x1) ) = x1

POL( false ) = 2

POL( MIN(x1) ) = max{0, 2x1 - 1}

POL( add(x1, x2) ) = 2x2 + 2


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(27) Obligation:

Q DP problem:
P is empty.
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) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES

(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) YES

(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) YES

(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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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) TransformationProof (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(nil, y1, y2) → IF(true, nil, y1, y2)
MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2) → MINS(add(x0, x1), y1, y2) → IF(false, add(x0, x1), y1, y2)

(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(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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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) DependencyGraphProof (EQUIVALENT transformation)

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

(52) 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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) 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.

(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
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)
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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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.

(55) 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))

(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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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.

(57) TransformationProof (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(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, 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, nil, y1, y2) → MINS(app(rm(head(nil), nil), y1), nil, app(y2, add(head(nil), nil)))

(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(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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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))) → IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(head(add(x0, x1)), 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(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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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))) → 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, 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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))) → IF2(true, nil, y1, y2) → MINS(app(nil, y1), nil, app(y2, add(head(nil), 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, 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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))) → IF2(true, add(x0, x1), y1, y2) → MINS(app(rm(x0, x1), y1), nil, app(y2, add(x0, 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, 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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))) → IF2(true, nil, y1, y2) → MINS(y1, nil, app(y2, add(head(nil), 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, 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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, 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) → IF2(false, nil, y1, y2) → MINS(nil, add(head(nil), y1), y2)

(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(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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) DependencyGraphProof (EQUIVALENT transformation)

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

(72) 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) 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.

(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:

head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
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)
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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) 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)

(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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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.

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

(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(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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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) → IF(false, add(z0, z1), z2, z3) → IF2(eq(head(add(z0, z1)), min(add(z0, z1))), add(z0, z1), z2, z3)

(80) 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) DependencyGraphProof (EQUIVALENT transformation)

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

(82) 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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) TransformationProof (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) → IF(false, add(z0, z1), z2, z3) → IF2(eq(z0, min(add(z0, z1))), add(z0, z1), z2, z3)

(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)
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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, 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) 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.

(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:

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))
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
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, 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.

(87) 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))

(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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
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)
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.

(89) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


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 Order [NEGPOLO,POLO] with Interpretation:
POL( MINS(x1, ..., x3) ) = x1 + x2

POL( app(x1, x2) ) = x1 + x2

POL( rm(x1, x2) ) = x2

POL( nil ) = 0

POL( add(x1, x2) ) = x1 + x2 + 1

POL( if_rm(x1, ..., x3) ) = x3

POL( eq(x1, x2) ) = 2

POL( true ) = 0

POL( IF2(x1, ..., x4) ) = x2 + x3

POL( min(x1) ) = max{0, x1 - 2}

POL( if_min(x1, x2) ) = x2

POL( le(x1, x2) ) = 0

POL( false ) = 0

POL( 0 ) = 1

POL( s(x1) ) = 0

POL( IF(x1, ..., x4) ) = x1 + x2 + x3


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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)
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))

(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(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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
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)
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) 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.

(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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
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.

(93) 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))

(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(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(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)
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.

(95) 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

(96) YES