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:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.

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

MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
MERGE2(.2(x, y), .2(u, v)) -> IF3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> 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:

MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
MERGE2(.2(x, y), .2(u, v)) -> IF3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

++12(.2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof

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

MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)

The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
Used argument filtering: MERGE2(x1, x2)  =  x2
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof

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

MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))

The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
Used argument filtering: MERGE2(x1, x2)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.