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:
f(n__a, n__b, X) → f(X, X, X)
c → a
c → b
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(n__a, n__b, X) → f(X, X, X)
c → a
c → b
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → 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:
C → B
C → A
ACTIVATE(n__a) → A
ACTIVATE(n__b) → B
F(n__a, n__b, X) → F(X, X, X)
The TRS R consists of the following rules:
f(n__a, n__b, X) → f(X, X, X)
c → a
c → b
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
C → B
C → A
ACTIVATE(n__a) → A
ACTIVATE(n__b) → B
F(n__a, n__b, X) → F(X, X, X)
The TRS R consists of the following rules:
f(n__a, n__b, X) → f(X, X, X)
c → a
c → b
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → 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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
F(n__a, n__b, X) → F(X, X, X)
The TRS R consists of the following rules:
f(n__a, n__b, X) → f(X, X, X)
c → a
c → b
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → 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:
F(n__a, n__b, X) → F(X, X, X)
The TRS R consists of the following rules:
f(n__a, n__b, X) → f(X, X, X)
c → a
c → b
a → n__a
b → n__b
activate(n__a) → a
activate(n__b) → b
activate(X) → X
s = F(c, c, X) evaluates to t =F(X, X, X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [X / c]
Rewriting sequence
F(c, c, c) → F(c, b, c)
with rule c → b at position [1] and matcher [ ]
F(c, b, c) → F(c, n__b, c)
with rule b → n__b at position [1] and matcher [ ]
F(c, n__b, c) → F(a, n__b, c)
with rule c → a at position [0] and matcher [ ]
F(a, n__b, c) → F(n__a, n__b, c)
with rule a → n__a at position [0] and matcher [ ]
F(n__a, n__b, c) → F(c, c, c)
with rule F(n__a, n__b, X) → F(X, X, 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.