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:

a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ RFCMatchBoundsTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(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(x1) → B(f(b(b(x1))))
A(x1) → B(b(x1))
B(d(b(x1))) → A(d(x1))
B(d(b(x1))) → D(x1)
A(d(x1)) → B(d(x1))
A(x1) → B(x1)
A(x1) → B(b(f(b(b(x1)))))
D(f(x1)) → D(x1)
D(f(x1)) → B(d(x1))
A(d(x1)) → B(c(b(d(x1))))
A(d(x1)) → D(b(c(b(d(x1)))))

The TRS R consists of the following rules:

a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof
  ↳ RFCMatchBoundsTRSProof

Q DP problem:
The TRS P consists of the following rules:

A(x1) → B(f(b(b(x1))))
A(x1) → B(b(x1))
B(d(b(x1))) → A(d(x1))
B(d(b(x1))) → D(x1)
A(d(x1)) → B(d(x1))
A(x1) → B(x1)
A(x1) → B(b(f(b(b(x1)))))
D(f(x1)) → D(x1)
D(f(x1)) → B(d(x1))
A(d(x1)) → B(c(b(d(x1))))
A(d(x1)) → D(b(c(b(d(x1)))))

The TRS R consists of the following rules:

a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(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 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
  ↳ RFCMatchBoundsTRSProof

Q DP problem:
The TRS P consists of the following rules:

A(x1) → B(b(x1))
B(d(b(x1))) → A(d(x1))
B(d(b(x1))) → D(x1)
A(d(x1)) → B(d(x1))
A(x1) → B(x1)
D(f(x1)) → D(x1)
D(f(x1)) → B(d(x1))

The TRS R consists of the following rules:

a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
Termination of the TRS R could be shown with a Match Bound [6,7] of 4. This implies Q-termination of R.
The following rules were used to construct the certificate:

a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

1, 2, 6, 3, 4, 5, 10, 9, 7, 8, 14, 13, 11, 12, 18, 15, 16, 17, 22, 21, 19, 20, 24, 25, 26, 23, 27, 31, 30, 28, 29, 33, 34, 35, 32, 42, 41, 39, 40, 48, 49, 50, 47, 53, 54, 55, 52, 57, 58, 59, 56, 63, 60, 61, 62, 67, 64, 65, 66

Node 1 is start node and node 2 is final node.

Those nodes are connect through the following edges: