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:

f3(x, y, z) -> g4(<=2(x, y), x, y, z)
g4(true, x, y, z) -> z
g4(false, x, y, z) -> f3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f3(x, y, z) -> g4(<=2(x, y), x, y, z)
g4(true, x, y, z) -> z
g4(false, x, y, z) -> f3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
p1(0) -> 0
p1(s1(x)) -> x

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

f3(x, y, z) -> g4(<=2(x, y), x, y, z)
g4(true, x, y, z) -> z
g4(false, x, y, z) -> f3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
p1(0) -> 0
p1(s1(x)) -> x

The set Q consists of the following terms:

f3(x0, x1, x2)
g4(true, x0, x1, x2)
g4(false, x0, x1, x2)
p1(0)
p1(s1(x0))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G4(false, x, y, z) -> P1(z)
G4(false, x, y, z) -> F3(p1(z), x, y)
G4(false, x, y, z) -> F3(p1(x), y, z)
G4(false, x, y, z) -> F3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
G4(false, x, y, z) -> P1(y)
F3(x, y, z) -> G4(<=2(x, y), x, y, z)
G4(false, x, y, z) -> F3(p1(y), z, x)
G4(false, x, y, z) -> P1(x)

The TRS R consists of the following rules:

f3(x, y, z) -> g4(<=2(x, y), x, y, z)
g4(true, x, y, z) -> z
g4(false, x, y, z) -> f3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
p1(0) -> 0
p1(s1(x)) -> x

The set Q consists of the following terms:

f3(x0, x1, x2)
g4(true, x0, x1, x2)
g4(false, x0, x1, x2)
p1(0)
p1(s1(x0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

G4(false, x, y, z) -> P1(z)
G4(false, x, y, z) -> F3(p1(z), x, y)
G4(false, x, y, z) -> F3(p1(x), y, z)
G4(false, x, y, z) -> F3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
G4(false, x, y, z) -> P1(y)
F3(x, y, z) -> G4(<=2(x, y), x, y, z)
G4(false, x, y, z) -> F3(p1(y), z, x)
G4(false, x, y, z) -> P1(x)

The TRS R consists of the following rules:

f3(x, y, z) -> g4(<=2(x, y), x, y, z)
g4(true, x, y, z) -> z
g4(false, x, y, z) -> f3(f3(p1(x), y, z), f3(p1(y), z, x), f3(p1(z), x, y))
p1(0) -> 0
p1(s1(x)) -> x

The set Q consists of the following terms:

f3(x0, x1, x2)
g4(true, x0, x1, x2)
g4(false, x0, x1, x2)
p1(0)
p1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 8 less nodes.