(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(X), s(Y)) → EQ(X, Y)
LE(s(X), s(Y)) → LE(X, Y)
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
MIN(cons(N, cons(M, L))) → LE(N, M)
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
REPLACE(N, M, cons(K, L)) → EQ(N, K)
IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
SELSORT(cons(N, L)) → EQ(N, min(cons(N, L)))
SELSORT(cons(N, L)) → MIN(cons(N, L))
IFSELSORT(true, cons(N, L)) → SELSORT(L)
IFSELSORT(false, cons(N, L)) → MIN(cons(N, L))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
IFSELSORT(false, cons(N, L)) → REPLACE(min(cons(N, L)), N, L)
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 6 less nodes.
(6) Complex Obligation (AND)
(7) 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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(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))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
(11) 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.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- LE(s(X), s(Y)) → LE(X, Y)
The graph contains the following edges 1 > 1, 2 > 2
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
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))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
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.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(IFMIN(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(le(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
none
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
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.
(21) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(22) TRUE
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(X), s(Y)) → EQ(X, Y)
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(24) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(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))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(26) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
(27) 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.
(28) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- EQ(s(X), s(Y)) → EQ(X, Y)
The graph contains the following edges 1 > 1, 2 > 2
(29) TRUE
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(31) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(33) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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.
(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:
- IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3
- REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4
(36) TRUE
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IFSELSORT(true, cons(N, L)) → SELSORT(L)
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
The TRS R consists of the following rules:
eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
The set Q consists of the following terms:
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
min(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(38) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IFSELSORT(true, cons(N, L)) → SELSORT(L)
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
The TRS R consists of the following rules:
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
eq(0, 0) → true
eq(0, s(Y)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
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(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(40) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
selsort(nil)
selsort(cons(x0, x1))
ifselsort(true, cons(x0, x1))
ifselsort(false, cons(x0, x1))
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IFSELSORT(true, cons(N, L)) → SELSORT(L)
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
The TRS R consists of the following rules:
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
eq(0, 0) → true
eq(0, s(Y)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
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(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
(42) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
IFSELSORT(true, cons(N, L)) → SELSORT(L)
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 1
POL(IFSELSORT(x1, x2)) = x2
POL(SELSORT(x1)) = x1
POL(cons(x1, x2)) = 1 + x2
POL(eq(x1, x2)) = x1
POL(false) = 0
POL(ifmin(x1, x2)) = 0
POL(ifrepl(x1, x2, x3, x4)) = x4
POL(le(x1, x2)) = 0
POL(min(x1)) = 0
POL(nil) = 0
POL(replace(x1, x2, x3)) = x3
POL(s(x1)) = x1
POL(true) = 0
The following usable rules [FROCOS05] were oriented:
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
replace(N, M, nil) → nil
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
The TRS R consists of the following rules:
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
eq(0, 0) → true
eq(0, s(Y)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
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(cons(0, nil))
min(cons(s(x0), nil))
min(cons(x0, cons(x1, x2)))
ifmin(true, cons(x0, cons(x1, x2)))
ifmin(false, cons(x0, cons(x1, x2)))
replace(x0, x1, nil)
replace(x0, x1, cons(x2, x3))
ifrepl(true, x0, x1, cons(x2, x3))
ifrepl(false, x0, x1, cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
(44) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(45) TRUE