(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) 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.
(9) 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.
(10) 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.
(11) 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.
(12) 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.