Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could not be shown:



Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

p(s(0), 0).
p(s(s(X)), s(s(Y))) :- p(s(X), s(Y)).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).

Queries:

plus(g,a,a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus_in: (b,f,f)
p_in: (b,f) (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(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:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(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:

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GG(x1, x2, x3)  =  U1_GG(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

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:

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GG(x1, x2, x3)  =  U1_GG(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

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

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

P_IN_GG(s, s) → P_IN_GG(s, s)

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:

P_IN_GG(s, s) → P_IN_GG(s, s)

The TRS R consists of the following rules:none


s = P_IN_GG(s, s) evaluates to t =P_IN_GG(s, s)

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 P_IN_GG(s, s) to P_IN_GG(s, s).





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

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

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ RuleRemovalProof
  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
U1_ga(p_out_gg) → p_out_ga(s)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))
U1_gg(p_out_gg) → p_out_gg

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

U1_ga(p_out_gg) → p_out_ga(s)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(PLUS_IN_GAA(x1)) = 2·x1   
POL(U1_ga(x1)) = x1   
POL(U1_gg(x1)) = x1   
POL(U2_GAA(x1)) = x1   
POL(p_in_ga(x1)) = x1   
POL(p_in_gg(x1, x2)) = x1 + x2   
POL(p_out_ga(x1)) = 2·x1   
POL(p_out_gg) = 1   
POL(s) = 0   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
QDP
                            ↳ RuleRemovalProof
  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))
U1_gg(p_out_gg) → p_out_gg

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

U1_gg(p_out_gg) → p_out_gg

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(PLUS_IN_GAA(x1)) = 2·x1   
POL(U1_ga(x1)) = x1   
POL(U1_gg(x1)) = 2·x1   
POL(U2_GAA(x1)) = x1   
POL(p_in_ga(x1)) = x1   
POL(p_in_gg(x1, x2)) = x1 + x2   
POL(p_out_ga(x1)) = 2·x1   
POL(p_out_gg) = 2   
POL(s) = 0   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
QDP
                                ↳ RuleRemovalProof
  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

p_in_ga(s) → U1_ga(p_in_gg(s, s))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(PLUS_IN_GAA(x1)) = 1 + 2·x1   
POL(U1_ga(x1)) = 2·x1   
POL(U1_gg(x1)) = x1   
POL(U2_GAA(x1)) = x1   
POL(p_in_ga(x1)) = 1 + x1   
POL(p_in_gg(x1, x2)) = 2·x1 + x2   
POL(p_out_ga(x1)) = 1 + 2·x1   
POL(s) = 0   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ RuleRemovalProof
QDP
                                    ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ RuleRemovalProof
                                  ↳ QDP
                                    ↳ UsableRulesProof
QDP
                                        ↳ QReductionProof
  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(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.

U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ RuleRemovalProof
                                  ↳ QDP
                                    ↳ UsableRulesProof
                                      ↳ QDP
                                        ↳ QReductionProof
QDP
                                            ↳ RuleRemovalProof
  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1   
POL(PLUS_IN_GAA(x1)) = 2·x1   
POL(U2_GAA(x1)) = x1   
POL(p_in_ga(x1)) = x1   
POL(p_out_ga(x1)) = 2·x1   
POL(s) = 2   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ RuleRemovalProof
                                  ↳ QDP
                                    ↳ UsableRulesProof
                                      ↳ QDP
                                        ↳ QReductionProof
                                          ↳ QDP
                                            ↳ RuleRemovalProof
QDP
                                                ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus_in: (b,f,f)
p_in: (b,f) (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(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:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(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:

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GG(x1, x2, x3)  =  U1_GG(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

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:

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GG(x1, x2, x3)  =  U1_GG(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

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

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s
P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

P_IN_GG(s, s) → P_IN_GG(s, s)

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:

P_IN_GG(s, s) → P_IN_GG(s, s)

The TRS R consists of the following rules:none


s = P_IN_GG(s, s) evaluates to t =P_IN_GG(s, s)

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 P_IN_GG(s, s) to P_IN_GG(s, s).





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

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

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
p_out_gg(x1, x2)  =  p_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x3)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ RuleRemovalProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
U1_ga(p_out_gg(s, s)) → p_out_ga(s, s)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))
U1_gg(p_out_gg(s, s)) → p_out_gg(s, s)

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

p_in_ga(s) → U1_ga(p_in_gg(s, s))
U1_gg(p_out_gg(s, s)) → p_out_gg(s, s)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(PLUS_IN_GAA(x1)) = 2 + 2·x1   
POL(U1_ga(x1)) = 2·x1   
POL(U1_gg(x1)) = 2·x1   
POL(U2_GAA(x1)) = x1   
POL(p_in_ga(x1)) = 2 + x1   
POL(p_in_gg(x1, x2)) = 2·x1 + 2·x2   
POL(p_out_ga(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(p_out_gg(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(s) = 0   



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
QDP
                            ↳ UsableRulesProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)
U1_ga(p_out_gg(s, s)) → p_out_ga(s, s)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ UsableRulesProof
QDP
                                ↳ QReductionProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(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.

U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
QDP
                                    ↳ RuleRemovalProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(PLUS_IN_GAA(x1)) = 2 + 2·x1   
POL(U2_GAA(x1)) = x1   
POL(p_in_ga(x1)) = 1 + 2·x1   
POL(p_out_ga(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(s) = 1   



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ RuleRemovalProof
QDP
                                        ↳ PisEmptyProof

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

p_in_ga(s) → p_out_ga(s, 0)

The set Q consists of the following terms:

p_in_ga(x0)

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