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(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(x1)))

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(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:

B(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(a(b(x1))) → A(c(a(x1)))
A(a(x1)) → B(a(x1))

The TRS R consists of the following rules:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(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:

B(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(a(b(x1))) → A(c(a(x1)))
A(a(x1)) → B(a(x1))

The TRS R consists of the following rules:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(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
          ↳ QDPOrderProof
  ↳ QTRS Reverse

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

B(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
A(a(x1)) → B(a(x1))

The TRS R consists of the following rules:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(x1)))

Q is empty.
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.


A(a(x1)) → B(a(x1))
The remaining pairs can at least be oriented weakly.

B(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = x1


POL( c(x1) ) = 0


POL( b(x1) ) = x1


POL( B(x1) ) = max{0, x1 - 1}


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(x1)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

B(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))

The TRS R consists of the following rules:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(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
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
QDP
  ↳ QTRS Reverse

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

A(a(x1)) → A(b(a(x1)))

The TRS R consists of the following rules:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(x1)))

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → a(b(a(x)))
b(a(b(x))) → a(c(a(x)))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
      ↳ RFCMatchBoundsTRSProof

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

a(a(x)) → a(b(a(x)))
b(a(b(x))) → a(c(a(x)))

Q is empty.

Termination of the TRS R could be shown with a Match Bound [6,7] of 5. This implies Q-termination of R.
The following rules were used to construct the certificate:

a(a(x)) → a(b(a(x)))
b(a(b(x))) → a(c(a(x)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 115, 114, 117, 116, 118, 119, 121, 120

Node 92 is start node and node 93 is final node.

Those nodes are connect through the following edges: