Left Termination of the query pattern ackermann_in_3(g, a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

ackermann(0, N, s(N)).
ackermann(s(M), 0, Val) :- ackermann(M, s(0), Val).
ackermann(s(M), s(N), Val) :- ','(ackermann(s(M), N, Val1), ackermann(M, Val1, Val)).

Queries:

ackermann(g,a,g).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x3)
ackermann_out(x1, x2, x3)  =  ackermann_out
U3(x1, x2, x3, x4)  =  U3(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x3)
ackermann_out(x1, x2, x3)  =  ackermann_out
U3(x1, x2, x3, x4)  =  U3(x4)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

ACKERMANN_IN(s(M), s(N), Val) → U21(M, N, Val, ackermann_in(s(M), N, Val1))
ACKERMANN_IN(s(M), s(N), Val) → ACKERMANN_IN(s(M), N, Val1)
ACKERMANN_IN(s(M), 0, Val) → U11(M, Val, ackermann_in(M, s(0), Val))
ACKERMANN_IN(s(M), 0, Val) → ACKERMANN_IN(M, s(0), Val)
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → U31(M, N, Val, ackermann_in(M, Val1, Val))
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → ACKERMANN_IN(M, Val1, Val)

The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x3)
ackermann_out(x1, x2, x3)  =  ackermann_out
U3(x1, x2, x3, x4)  =  U3(x4)
U31(x1, x2, x3, x4)  =  U31(x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
ACKERMANN_IN(x1, x2, x3)  =  ACKERMANN_IN(x1)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M), s(N), Val) → U21(M, N, Val, ackermann_in(s(M), N, Val1))
ACKERMANN_IN(s(M), s(N), Val) → ACKERMANN_IN(s(M), N, Val1)
ACKERMANN_IN(s(M), 0, Val) → U11(M, Val, ackermann_in(M, s(0), Val))
ACKERMANN_IN(s(M), 0, Val) → ACKERMANN_IN(M, s(0), Val)
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → U31(M, N, Val, ackermann_in(M, Val1, Val))
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → ACKERMANN_IN(M, Val1, Val)

The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x3)
ackermann_out(x1, x2, x3)  =  ackermann_out
U3(x1, x2, x3, x4)  =  U3(x4)
U31(x1, x2, x3, x4)  =  U31(x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
ACKERMANN_IN(x1, x2, x3)  =  ACKERMANN_IN(x1)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M), s(N), Val) → U21(M, N, Val, ackermann_in(s(M), N, Val1))
ACKERMANN_IN(s(M), s(N), Val) → ACKERMANN_IN(s(M), N, Val1)
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → ACKERMANN_IN(M, Val1, Val)
ACKERMANN_IN(s(M), 0, Val) → ACKERMANN_IN(M, s(0), Val)

The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x3)
ackermann_out(x1, x2, x3)  =  ackermann_out
U3(x1, x2, x3, x4)  =  U3(x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
ACKERMANN_IN(x1, x2, x3)  =  ACKERMANN_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(M)
ACKERMANN_IN(s(M)) → U21(M, ackermann_in(s(M)))
U21(M, ackermann_out) → ACKERMANN_IN(M)
ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:

ackermann_in(s(M)) → U2(M, ackermann_in(s(M)))
ackermann_in(s(M)) → U1(ackermann_in(M))
ackermann_in(0) → ackermann_out
U1(ackermann_out) → ackermann_out
U2(M, ackermann_out) → U3(ackermann_in(M))
U3(ackermann_out) → ackermann_out

The set Q consists of the following terms:

ackermann_in(x0)
U1(x0)
U2(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACKERMANN_IN(s(M)) → ACKERMANN_IN(M)
U21(M, ackermann_out) → ACKERMANN_IN(M)
The remaining pairs can at least be oriented weakly.

ACKERMANN_IN(s(M)) → U21(M, ackermann_in(s(M)))
ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))
Used ordering: Polynomial interpretation [25]:

POL(0) = 1   
POL(ACKERMANN_IN(x1)) = x1   
POL(U1(x1)) = 0   
POL(U2(x1, x2)) = 0   
POL(U21(x1, x2)) = 1 + x1   
POL(U3(x1)) = 0   
POL(ackermann_in(x1)) = 1 + x1   
POL(ackermann_out) = 1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented: none



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M)) → U21(M, ackermann_in(s(M)))
ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:

ackermann_in(s(M)) → U2(M, ackermann_in(s(M)))
ackermann_in(s(M)) → U1(ackermann_in(M))
ackermann_in(0) → ackermann_out
U1(ackermann_out) → ackermann_out
U2(M, ackermann_out) → U3(ackermann_in(M))
U3(ackermann_out) → ackermann_out

The set Q consists of the following terms:

ackermann_in(x0)
U1(x0)
U2(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:

ackermann_in(s(M)) → U2(M, ackermann_in(s(M)))
ackermann_in(s(M)) → U1(ackermann_in(M))
ackermann_in(0) → ackermann_out
U1(ackermann_out) → ackermann_out
U2(M, ackermann_out) → U3(ackermann_in(M))
U3(ackermann_out) → ackermann_out

The set Q consists of the following terms:

ackermann_in(x0)
U1(x0)
U2(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

R is empty.
The set Q consists of the following terms:

ackermann_in(x0)
U1(x0)
U2(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ackermann_in(x0)
U1(x0)
U2(x0, x1)
U3(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:none


s = ACKERMANN_IN(s(M)) evaluates to t =ACKERMANN_IN(s(M))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ACKERMANN_IN(s(M)) to ACKERMANN_IN(s(M)).




We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x1, x3)
ackermann_out(x1, x2, x3)  =  ackermann_out(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x1, x3)
ackermann_out(x1, x2, x3)  =  ackermann_out(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

ACKERMANN_IN(s(M), s(N), Val) → U21(M, N, Val, ackermann_in(s(M), N, Val1))
ACKERMANN_IN(s(M), s(N), Val) → ACKERMANN_IN(s(M), N, Val1)
ACKERMANN_IN(s(M), 0, Val) → U11(M, Val, ackermann_in(M, s(0), Val))
ACKERMANN_IN(s(M), 0, Val) → ACKERMANN_IN(M, s(0), Val)
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → U31(M, N, Val, ackermann_in(M, Val1, Val))
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → ACKERMANN_IN(M, Val1, Val)

The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x1, x3)
ackermann_out(x1, x2, x3)  =  ackermann_out(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
U31(x1, x2, x3, x4)  =  U31(x1, x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
ACKERMANN_IN(x1, x2, x3)  =  ACKERMANN_IN(x1)
U11(x1, x2, x3)  =  U11(x1, x3)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

ACKERMANN_IN(s(M), s(N), Val) → U21(M, N, Val, ackermann_in(s(M), N, Val1))
ACKERMANN_IN(s(M), s(N), Val) → ACKERMANN_IN(s(M), N, Val1)
ACKERMANN_IN(s(M), 0, Val) → U11(M, Val, ackermann_in(M, s(0), Val))
ACKERMANN_IN(s(M), 0, Val) → ACKERMANN_IN(M, s(0), Val)
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → U31(M, N, Val, ackermann_in(M, Val1, Val))
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → ACKERMANN_IN(M, Val1, Val)

The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x1, x3)
ackermann_out(x1, x2, x3)  =  ackermann_out(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
U31(x1, x2, x3, x4)  =  U31(x1, x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
ACKERMANN_IN(x1, x2, x3)  =  ACKERMANN_IN(x1)
U11(x1, x2, x3)  =  U11(x1, x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof

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

ACKERMANN_IN(s(M), s(N), Val) → U21(M, N, Val, ackermann_in(s(M), N, Val1))
ACKERMANN_IN(s(M), s(N), Val) → ACKERMANN_IN(s(M), N, Val1)
U21(M, N, Val, ackermann_out(s(M), N, Val1)) → ACKERMANN_IN(M, Val1, Val)
ACKERMANN_IN(s(M), 0, Val) → ACKERMANN_IN(M, s(0), Val)

The TRS R consists of the following rules:

ackermann_in(s(M), s(N), Val) → U2(M, N, Val, ackermann_in(s(M), N, Val1))
ackermann_in(s(M), 0, Val) → U1(M, Val, ackermann_in(M, s(0), Val))
ackermann_in(0, N, s(N)) → ackermann_out(0, N, s(N))
U1(M, Val, ackermann_out(M, s(0), Val)) → ackermann_out(s(M), 0, Val)
U2(M, N, Val, ackermann_out(s(M), N, Val1)) → U3(M, N, Val, ackermann_in(M, Val1, Val))
U3(M, N, Val, ackermann_out(M, Val1, Val)) → ackermann_out(s(M), s(N), Val)

The argument filtering Pi contains the following mapping:
ackermann_in(x1, x2, x3)  =  ackermann_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
0  =  0
U1(x1, x2, x3)  =  U1(x1, x3)
ackermann_out(x1, x2, x3)  =  ackermann_out(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
ACKERMANN_IN(x1, x2, x3)  =  ACKERMANN_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(M)
ACKERMANN_IN(s(M)) → U21(M, ackermann_in(s(M)))
U21(M, ackermann_out(s(M))) → ACKERMANN_IN(M)
ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:

ackermann_in(s(M)) → U2(M, ackermann_in(s(M)))
ackermann_in(s(M)) → U1(M, ackermann_in(M))
ackermann_in(0) → ackermann_out(0)
U1(M, ackermann_out(M)) → ackermann_out(s(M))
U2(M, ackermann_out(s(M))) → U3(M, ackermann_in(M))
U3(M, ackermann_out(M)) → ackermann_out(s(M))

The set Q consists of the following terms:

ackermann_in(x0)
U1(x0, x1)
U2(x0, x1)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACKERMANN_IN(s(M)) → ACKERMANN_IN(M)
U21(M, ackermann_out(s(M))) → ACKERMANN_IN(M)
The remaining pairs can at least be oriented weakly.

ACKERMANN_IN(s(M)) → U21(M, ackermann_in(s(M)))
ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))
Used ordering: Polynomial interpretation [25]:

POL(0) = 1   
POL(ACKERMANN_IN(x1)) = x1   
POL(U1(x1, x2)) = 0   
POL(U2(x1, x2)) = 0   
POL(U21(x1, x2)) = 1 + x1   
POL(U3(x1, x2)) = 0   
POL(ackermann_in(x1)) = 1 + x1   
POL(ackermann_out(x1)) = 1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented: none



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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

ACKERMANN_IN(s(M)) → U21(M, ackermann_in(s(M)))
ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:

ackermann_in(s(M)) → U2(M, ackermann_in(s(M)))
ackermann_in(s(M)) → U1(M, ackermann_in(M))
ackermann_in(0) → ackermann_out(0)
U1(M, ackermann_out(M)) → ackermann_out(s(M))
U2(M, ackermann_out(s(M))) → U3(M, ackermann_in(M))
U3(M, ackermann_out(M)) → ackermann_out(s(M))

The set Q consists of the following terms:

ackermann_in(x0)
U1(x0, x1)
U2(x0, x1)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:

ackermann_in(s(M)) → U2(M, ackermann_in(s(M)))
ackermann_in(s(M)) → U1(M, ackermann_in(M))
ackermann_in(0) → ackermann_out(0)
U1(M, ackermann_out(M)) → ackermann_out(s(M))
U2(M, ackermann_out(s(M))) → U3(M, ackermann_in(M))
U3(M, ackermann_out(M)) → ackermann_out(s(M))

The set Q consists of the following terms:

ackermann_in(x0)
U1(x0, x1)
U2(x0, x1)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

R is empty.
The set Q consists of the following terms:

ackermann_in(x0)
U1(x0, x1)
U2(x0, x1)
U3(x0, x1)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ackermann_in(x0)
U1(x0, x1)
U2(x0, x1)
U3(x0, x1)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ NonTerminationProof

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

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

ACKERMANN_IN(s(M)) → ACKERMANN_IN(s(M))

The TRS R consists of the following rules:none


s = ACKERMANN_IN(s(M)) evaluates to t =ACKERMANN_IN(s(M))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ACKERMANN_IN(s(M)) to ACKERMANN_IN(s(M)).