Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(c(b(x1))))
A(b(x1)) → C(b(x1))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(c(b(x1))))
A(b(x1)) → C(b(x1))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(c(b(x1))))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(a(c(b(x1)))) at position [0] we obtained the following new rules:
A(b(y0)) → A(c(b(y0)))
A(b(x0)) → A(a(c(a(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → A(c(b(x1)))
A(b(x0)) → A(a(c(a(x0))))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(c(b(x1))) at position [0] we obtained the following new rules:
A(b(x0)) → A(c(a(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x0)) → A(c(a(x0)))
B(x1) → A(x1)
A(b(x0)) → A(a(c(a(x0))))
A(b(x1)) → B(a(a(c(b(x1)))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
A(b(x0)) → A(c(a(x0)))
B(x1) → A(x1)
A(b(x0)) → A(a(c(a(x0))))
A(b(x1)) → B(a(a(c(b(x1)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
A(b(x0)) → A(c(a(x0)))
B(x1) → A(x1)
A(b(x0)) → A(a(c(a(x0))))
A(b(x1)) → B(a(a(c(b(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(A(x)) → A1(a(B(x)))
B1(a(x)) → A1(a(b(x)))
B1(a(x)) → A1(b(x))
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(x)) → B2(x)
B1(A(x)) → C(a(A(x)))
B1(A(x)) → C(a(a(B(x))))
B1(A(x)) → C(A(x))
B1(A(x)) → B1(c(a(a(B(x)))))
B1(A(x)) → A1(c(a(A(x))))
B1(A(x)) → A1(A(x))
B1(A(x)) → A1(B(x))
B1(x) → A1(x)
B1(a(x)) → C(a(a(b(x))))
B1(A(x)) → A1(c(A(x)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(A(x)) → A1(a(B(x)))
B1(a(x)) → A1(a(b(x)))
B1(a(x)) → A1(b(x))
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(x)) → B2(x)
B1(A(x)) → C(a(A(x)))
B1(A(x)) → C(a(a(B(x))))
B1(A(x)) → C(A(x))
B1(A(x)) → B1(c(a(a(B(x)))))
B1(A(x)) → A1(c(a(A(x))))
B1(A(x)) → A1(A(x))
B1(A(x)) → A1(B(x))
B1(x) → A1(x)
B1(a(x)) → C(a(a(b(x))))
B1(A(x)) → A1(c(A(x)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 13 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(A(x)) → B1(c(a(a(B(x)))))
B1(a(x)) → B1(c(a(a(b(x)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(x)) → B1(c(a(a(B(x))))) at position [0] we obtained the following new rules:
B1(A(x0)) → B1(c(a(a(A(x0)))))
B1(A(y0)) → B1(c(a(B(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(A(y0)) → B1(c(a(B(y0))))
B1(A(x0)) → B1(c(a(a(A(x0)))))
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(x0)) → B1(c(a(a(A(x0))))) at position [0] we obtained the following new rules:
B1(A(y0)) → B1(c(a(A(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(A(y0)) → B1(c(a(B(y0))))
B1(a(x)) → B1(x)
B1(A(y0)) → B1(c(a(A(y0))))
B1(a(x)) → B1(c(a(a(b(x)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(y0)) → B1(c(a(B(y0)))) at position [0] we obtained the following new rules:
B1(A(y0)) → B1(c(B(y0)))
B1(A(x0)) → B1(c(a(A(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(A(y0)) → B1(c(a(A(y0))))
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(y0)) → B1(c(B(y0)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(y0)) → B1(c(a(A(y0)))) at position [0] we obtained the following new rules:
B1(A(y0)) → B1(c(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(A(y0)) → B1(c(A(y0)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(y0)) → B1(c(B(y0)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
B1(A(y0)) → B1(c(B(y0)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(A(y0)) → B1(c(B(y0))) at position [0] we obtained the following new rules:
B1(A(x0)) → B1(c(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(A(x0)) → B1(c(A(x0)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(a(x)) → B1(c(a(a(b(x)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(A(x)) → B1(c(a(a(B(x)))))
B1(a(x)) → B1(c(a(a(b(x)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
s = B1(c(a(a(b(A(x')))))) evaluates to t =B1(c(a(a(b(A(x'))))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
B1(c(a(a(b(A(x')))))) → B1(c(a(b(A(x')))))
with rule a(x) → x at position [0,0] and matcher [x / a(b(A(x')))]
B1(c(a(b(A(x'))))) → B1(c(b(A(x'))))
with rule a(x) → x at position [0,0] and matcher [x / b(A(x'))]
B1(c(b(A(x')))) → B1(c(a(c(a(A(x'))))))
with rule b(A(x'')) → a(c(a(A(x'')))) at position [0,0] and matcher [x'' / x']
B1(c(a(c(a(A(x')))))) → B1(c(c(a(A(x')))))
with rule a(x'') → x'' at position [0,0] and matcher [x'' / c(a(A(x')))]
B1(c(c(a(A(x'))))) → B1(a(A(x')))
with rule c(c(x'')) → x'' at position [0] and matcher [x'' / a(A(x'))]
B1(a(A(x'))) → B1(c(a(a(b(A(x'))))))
with rule B1(a(x)) → B1(c(a(a(b(x)))))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(a(a(c(b(x)))))
b(x) → a(x)
c(c(x)) → x
A(b(x)) → A(c(a(x)))
B(x) → A(x)
A(b(x)) → A(a(c(a(x))))
A(b(x)) → B(a(a(c(b(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ NonTerminationProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(x)) → b(a(a(c(b(x)))))
b(x) → a(x)
c(c(x)) → x
A(b(x)) → A(c(a(x)))
B(x) → A(x)
A(b(x)) → A(a(c(a(x))))
A(b(x)) → B(a(a(c(b(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
b(A(x)) → a(c(A(x)))
B(x) → A(x)
b(A(x)) → a(c(a(A(x))))
b(A(x)) → b(c(a(a(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(a(a(c(b(x)))))
b(x) → a(x)
c(c(x)) → x
A(b(x)) → A(c(a(x)))
B(x) → A(x)
A(b(x)) → A(a(c(a(x))))
A(b(x)) → B(a(a(c(b(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ NonTerminationProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(x)) → b(a(a(c(b(x)))))
b(x) → a(x)
c(c(x)) → x
A(b(x)) → A(c(a(x)))
B(x) → A(x)
A(b(x)) → A(a(c(a(x))))
A(b(x)) → B(a(a(c(b(x)))))
Q is empty.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
A(b(x1)) → A(c(b(x1)))
A(b(x1)) → A(a(c(b(x1))))
B(x1) → A(x1)
A(b(x1)) → B(a(a(c(b(x1)))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
s = A(a(c(b(a(a(c(b(x1')))))))) evaluates to t =A(a(c(b(a(a(c(b(x1'))))))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
A(a(c(b(a(a(c(b(x1')))))))) → A(a(c(b(a(c(b(x1')))))))
with rule a(x1) → x1 at position [0,0,0,0] and matcher [x1 / a(c(b(x1')))]
A(a(c(b(a(c(b(x1'))))))) → A(a(c(b(c(b(x1'))))))
with rule a(x1) → x1 at position [0,0,0,0] and matcher [x1 / c(b(x1'))]
A(a(c(b(c(b(x1')))))) → A(a(c(a(c(b(x1'))))))
with rule b(x1) → a(x1) at position [0,0,0] and matcher [x1 / c(b(x1'))]
A(a(c(a(c(b(x1')))))) → A(a(c(c(b(x1')))))
with rule a(x1) → x1 at position [0,0,0] and matcher [x1 / c(b(x1'))]
A(a(c(c(b(x1'))))) → A(a(b(x1')))
with rule c(c(x1)) → x1 at position [0,0] and matcher [x1 / b(x1')]
A(a(b(x1'))) → A(b(a(a(c(b(x1'))))))
with rule a(b(x1'')) → b(a(a(c(b(x1''))))) at position [0] and matcher [x1'' / x1']
A(b(a(a(c(b(x1')))))) → A(a(c(b(a(a(c(b(x1'))))))))
with rule A(b(x1)) → A(a(c(b(x1))))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(a(a(c(b(x1)))))
b(x1) → a(x1)
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → b(c(a(a(b(x)))))
b(x) → a(x)
c(c(x)) → x
Q is empty.