Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INTLIST(x) → IF_INTLIST(empty(x), x)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INTLIST(false, x) → TAIL(x)
IF_INTLIST(false, x) → INTLIST(tail(x))
IF2(false, x, y) → P(x)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF_INTLIST(false, x) → HEAD(x)
IF2(false, x, y) → INTLIST(int(p(x), p(y)))
P(s(s(x))) → P(s(x))
INT(x, y) → ZERO(y)
INT(x, y) → ZERO(x)
IF2(false, x, y) → P(y)
IF2(false, x, y) → INT(p(x), p(y))
INTLIST(x) → EMPTY(x)
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INTLIST(x) → IF_INTLIST(empty(x), x)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INTLIST(false, x) → TAIL(x)
IF_INTLIST(false, x) → INTLIST(tail(x))
IF2(false, x, y) → P(x)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF_INTLIST(false, x) → HEAD(x)
IF2(false, x, y) → INTLIST(int(p(x), p(y)))
P(s(s(x))) → P(s(x))
INT(x, y) → ZERO(y)
INT(x, y) → ZERO(x)
IF2(false, x, y) → P(y)
IF2(false, x, y) → INT(p(x), p(y))
INTLIST(x) → EMPTY(x)
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 8 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(x) → IF_INTLIST(empty(x), x)
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(x) → IF_INTLIST(empty(x), x)
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
empty(nil) → true
empty(cons(x, y)) → false
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(x) → IF_INTLIST(empty(x), x)
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
empty(nil) → true
empty(cons(x, y)) → false
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule INTLIST(x) → IF_INTLIST(empty(x), x) at position [0] we obtained the following new rules:
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
INTLIST(nil) → IF_INTLIST(true, nil)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
INTLIST(nil) → IF_INTLIST(true, nil)
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
empty(nil) → true
empty(cons(x, y)) → false
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
empty(nil) → true
empty(cons(x, y)) → false
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
empty(nil)
empty(cons(x0, x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
IF_INTLIST(false, x) → INTLIST(tail(x))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
The set Q consists of the following terms:
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF_INTLIST(false, x) → INTLIST(tail(x)) at position [0] we obtained the following new rules:
IF_INTLIST(false, cons(x0, x1)) → INTLIST(x1)
IF_INTLIST(false, nil) → INTLIST(nil)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_INTLIST(false, cons(x0, x1)) → INTLIST(x1)
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
IF_INTLIST(false, nil) → INTLIST(nil)
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
The set Q consists of the following terms:
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_INTLIST(false, cons(x0, x1)) → INTLIST(x1)
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
The TRS R consists of the following rules:
tail(nil) → nil
tail(cons(x, y)) → y
The set Q consists of the following terms:
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_INTLIST(false, cons(x0, x1)) → INTLIST(x1)
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
R is empty.
The set Q consists of the following terms:
tail(nil)
tail(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
tail(nil)
tail(cons(x0, x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_INTLIST(false, cons(x0, x1)) → INTLIST(x1)
INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- INTLIST(cons(x0, x1)) → IF_INTLIST(false, cons(x0, x1))
The graph contains the following edges 1 >= 2
- IF_INTLIST(false, cons(x0, x1)) → INTLIST(x1)
The graph contains the following edges 2 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
R is empty.
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- P(s(s(x))) → P(s(x))
The graph contains the following edges 1 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF2(false, x, y) → INT(p(x), p(y))
The TRS R consists of the following rules:
empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF2(false, x, y) → INT(p(x), p(y))
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF1(false, x, y) → INT(s(0), y)
IF_INT(false, b, x, y) → IF2(b, x, y)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF2(false, x, y) → INT(p(x), p(y))
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule INT(x, y) → IF_INT(zero(x), zero(y), x, y) at position [0] we obtained the following new rules:
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
IF_INT(true, b, x, y) → IF1(b, x, y)
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF2(false, x, y) → INT(p(x), p(y))
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_INT(true, b, x, y) → IF1(b, x, y) we obtained the following new rules:
IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF1(false, x, y) → INT(s(0), y)
IF_INT(false, b, x, y) → IF2(b, x, y)
IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0)
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF2(false, x, y) → INT(p(x), p(y))
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF1(false, x, y) → INT(s(0), y) we obtained the following new rules:
IF1(false, 0, z1) → INT(s(0), z1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, b, x, y) → IF2(b, x, y)
IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0)
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF1(false, 0, z1) → INT(s(0), z1)
IF2(false, x, y) → INT(p(x), p(y))
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF2(false, x, y) → INT(p(x), p(y)) at position [0] we obtained the following new rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF2(false, 0, y1) → INT(0, p(y1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF_INT(false, b, x, y) → IF2(b, x, y)
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0)
IF2(false, 0, y1) → INT(0, p(y1))
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_INT(false, b, x, y) → IF2(b, x, y) we obtained the following new rules:
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0)
IF2(false, 0, y1) → INT(0, p(y1))
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0)
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_INT(true, y_0, 0, z0) → IF1(y_0, 0, z0) we obtained the following new rules:
IF_INT(true, false, 0, x1) → IF1(false, 0, x1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF_INT(true, false, 0, x1) → IF1(false, 0, x1)
INT(0, y1) → IF_INT(true, zero(y1), 0, y1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule INT(0, y1) → IF_INT(true, zero(y1), 0, y1) at position [1] we obtained the following new rules:
INT(0, 0) → IF_INT(true, true, 0, 0)
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
INT(0, 0) → IF_INT(true, true, 0, 0)
IF_INT(true, false, 0, x1) → IF1(false, 0, x1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(0), y1) → INT(0, p(y1))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF_INT(true, false, 0, x1) → IF1(false, 0, x1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF2(false, s(0), y1) → INT(0, p(y1)) at position [1] we obtained the following new rules:
IF2(false, s(0), s(0)) → INT(0, 0)
IF2(false, s(0), 0) → INT(0, 0)
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), 0) → INT(0, 0)
IF2(false, s(0), s(0)) → INT(0, 0)
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF_INT(true, false, 0, x1) → IF1(false, 0, x1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF_INT(true, false, 0, x1) → IF1(false, 0, x1)
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_INT(true, false, 0, x1) → IF1(false, 0, x1) we obtained the following new rules:
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
IF1(false, 0, z1) → INT(s(0), z1)
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF1(false, 0, z1) → INT(s(0), z1) we obtained the following new rules:
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1)
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_INT(false, y_0, s(z0), z1) → IF2(y_0, s(z0), z1) we obtained the following new rules:
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1)
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule INT(s(x0), y1) → IF_INT(false, zero(y1), s(x0), y1) at position [1] we obtained the following new rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
INT(s(y0), 0) → IF_INT(false, true, s(y0), 0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(s(y0), 0) → IF_INT(false, true, s(y0), 0)
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
p(0) → 0
The set Q consists of the following terms:
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
zero(0)
zero(s(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1))
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
p(0) → 0
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF2(false, s(s(x0)), y1) → INT(s(p(s(x0))), p(y1)) at position [1] we obtained the following new rules:
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF2(false, s(s(y0)), s(0)) → INT(s(p(s(y0))), 0)
IF2(false, s(s(y0)), 0) → INT(s(p(s(y0))), 0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF2(false, s(s(y0)), s(0)) → INT(s(p(s(y0))), 0)
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF2(false, s(s(y0)), 0) → INT(s(p(s(y0))), 0)
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
p(0) → 0
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
p(0) → 0
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2)
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_INT(false, false, s(s(y_0)), x2) → IF2(false, s(s(y_0)), x2) we obtained the following new rules:
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0))
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule INT(s(y0), s(x0)) → IF_INT(false, false, s(y0), s(x0)) we obtained the following new rules:
INT(s(s(y_0)), s(x1)) → IF_INT(false, false, s(s(y_0)), s(x1))
INT(s(0), s(s(y_0))) → IF_INT(false, false, s(0), s(s(y_0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
INT(s(s(y_0)), s(x1)) → IF_INT(false, false, s(s(y_0)), s(x1))
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
INT(s(0), s(s(y_0))) → IF_INT(false, false, s(0), s(s(y_0)))
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ ForwardInstantiation
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
INT(s(0), s(s(y_0))) → IF_INT(false, false, s(0), s(s(y_0)))
IF1(false, 0, s(z0)) → INT(s(0), s(z0))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF1(false, 0, s(z0)) → INT(s(0), s(z0)) we obtained the following new rules:
IF1(false, 0, s(s(y_0))) → INT(s(0), s(s(y_0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF1(false, 0, s(s(y_0))) → INT(s(0), s(s(y_0)))
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
INT(s(0), s(s(y_0))) → IF_INT(false, false, s(0), s(s(y_0)))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
IF2(false, s(0), s(s(x0))) → INT(0, s(p(s(x0))))
The remaining pairs can at least be oriented weakly.
IF1(false, 0, s(s(y_0))) → INT(s(0), s(s(y_0)))
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
INT(s(0), s(s(y_0))) → IF_INT(false, false, s(0), s(s(y_0)))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
Tuple symbols:
| M( IF2(x1, ..., x3) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| M( INT(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
| M( IF1(x1, ..., x3) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| M( IF_INT(x1, ..., x4) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 | + | | · | x4 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF1(false, 0, s(s(y_0))) → INT(s(0), s(s(y_0)))
IF_INT(false, false, s(0), s(s(y_0))) → IF2(false, s(0), s(s(y_0)))
INT(0, s(x0)) → IF_INT(true, false, 0, s(x0))
INT(s(0), s(s(y_0))) → IF_INT(false, false, s(0), s(s(y_0)))
IF_INT(true, false, 0, s(z0)) → IF1(false, 0, s(z0))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 5 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
INT(s(s(y_0)), s(x1)) → IF_INT(false, false, s(s(y_0)), s(x1))
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
p(s(0)) → 0
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 2
POL(IF2(x1, x2, x3)) = 2·x1 + x2 + x3
POL(IF_INT(x1, x2, x3, x4)) = x1 + 2·x2 + x3 + x4
POL(INT(x1, x2)) = x1 + x2
POL(false) = 0
POL(p(x1)) = x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(s(y_0)), s(x1)) → IF_INT(false, false, s(s(y_0)), s(x1))
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
IF2(false, s(s(y0)), s(s(x0))) → INT(s(p(s(y0))), s(p(s(x0))))
The remaining pairs can at least be oriented weakly.
INT(s(s(y_0)), s(x1)) → IF_INT(false, false, s(s(y_0)), s(x1))
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
Tuple symbols:
| M( IF2(x1, ..., x3) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| M( INT(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
| M( IF_INT(x1, ..., x4) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 | + | | · | x4 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
p(s(s(x))) → s(p(s(x)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
INT(s(s(y_0)), s(x1)) → IF_INT(false, false, s(s(y_0)), s(x1))
IF_INT(false, false, s(s(x0)), s(z1)) → IF2(false, s(s(x0)), s(z1))
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(0)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.