MAYBE
Termination w.r.t. Q proof of ../tpdb/TRS/secret06/matchbox/gen-28.trs
Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)
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:
B(y, z) → C(y, z, z)
C(f(z), f(c(a, x, a)), y) → B(x, z)
C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
B(y, z) → C(c(y, z, z), a, a)
C(f(z), f(c(a, x, a)), y) → C(z, y, a)
The TRS R consists of the following rules:
b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)
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:
B(y, z) → C(y, z, z)
C(f(z), f(c(a, x, a)), y) → B(x, z)
C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
B(y, z) → C(c(y, z, z), a, a)
C(f(z), f(c(a, x, a)), y) → C(z, y, a)
The TRS R consists of the following rules:
b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)
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
Q DP problem:
The TRS P consists of the following rules:
B(y, z) → C(y, z, z)
C(f(z), f(c(a, x, a)), y) → B(x, z)
C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
C(f(z), f(c(a, x, a)), y) → C(z, y, a)
The TRS R consists of the following rules:
b(y, z) → f(c(c(y, z, z), a, a))
b(b(z, y), a) → z
c(f(z), f(c(a, x, a)), y) → c(f(b(x, z)), c(z, y, a), a)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.