Left Termination of the query pattern convert_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

convert([], B, 0).
convert(.(0, XS), B, X) :- ','(convert(XS, B, Y), times(Y, B, X)).
convert(.(s(Y), XS), B, s(X)) :- convert(.(Y, XS), B, X).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- plus(X, Y, Z).
times(0, Y, 0).
times(s(X), Y, Z) :- ','(times(X, Y, U), plus(Y, U, Z)).

Queries:

convert(g,g,a).

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:

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)


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

CONVERT_IN(.(s(Y), XS), B, s(X)) → U31(Y, XS, B, X, convert_in(.(Y, XS), B, X))
CONVERT_IN(.(s(Y), XS), B, s(X)) → CONVERT_IN(.(Y, XS), B, X)
CONVERT_IN(.(0, XS), B, X) → U11(XS, B, X, convert_in(XS, B, Y))
CONVERT_IN(.(0, XS), B, X) → CONVERT_IN(XS, B, Y)
U11(XS, B, X, convert_out(XS, B, Y)) → U21(XS, B, X, times_in(Y, B, X))
U11(XS, B, X, convert_out(XS, B, Y)) → TIMES_IN(Y, B, X)
TIMES_IN(s(X), Y, Z) → U51(X, Y, Z, times_in(X, Y, U))
TIMES_IN(s(X), Y, Z) → TIMES_IN(X, Y, U)
U51(X, Y, Z, times_out(X, Y, U)) → U61(X, Y, Z, plus_in(Y, U, Z))
U51(X, Y, Z, times_out(X, Y, U)) → PLUS_IN(Y, U, Z)
PLUS_IN(s(X), Y, s(Z)) → U41(X, Y, Z, plus_in(X, Y, Z))
PLUS_IN(s(X), Y, s(Z)) → PLUS_IN(X, Y, Z)

The TRS R consists of the following rules:

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)
U51(x1, x2, x3, x4)  =  U51(x2, x4)
TIMES_IN(x1, x2, x3)  =  TIMES_IN(x1, x2)
U41(x1, x2, x3, x4)  =  U41(x4)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
CONVERT_IN(x1, x2, x3)  =  CONVERT_IN(x1, x2)
U21(x1, x2, x3, x4)  =  U21(x4)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1, x2)
U61(x1, x2, x3, x4)  =  U61(x4)
U11(x1, x2, x3, x4)  =  U11(x2, x4)

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

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

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

CONVERT_IN(.(s(Y), XS), B, s(X)) → U31(Y, XS, B, X, convert_in(.(Y, XS), B, X))
CONVERT_IN(.(s(Y), XS), B, s(X)) → CONVERT_IN(.(Y, XS), B, X)
CONVERT_IN(.(0, XS), B, X) → U11(XS, B, X, convert_in(XS, B, Y))
CONVERT_IN(.(0, XS), B, X) → CONVERT_IN(XS, B, Y)
U11(XS, B, X, convert_out(XS, B, Y)) → U21(XS, B, X, times_in(Y, B, X))
U11(XS, B, X, convert_out(XS, B, Y)) → TIMES_IN(Y, B, X)
TIMES_IN(s(X), Y, Z) → U51(X, Y, Z, times_in(X, Y, U))
TIMES_IN(s(X), Y, Z) → TIMES_IN(X, Y, U)
U51(X, Y, Z, times_out(X, Y, U)) → U61(X, Y, Z, plus_in(Y, U, Z))
U51(X, Y, Z, times_out(X, Y, U)) → PLUS_IN(Y, U, Z)
PLUS_IN(s(X), Y, s(Z)) → U41(X, Y, Z, plus_in(X, Y, Z))
PLUS_IN(s(X), Y, s(Z)) → PLUS_IN(X, Y, Z)

The TRS R consists of the following rules:

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)
U51(x1, x2, x3, x4)  =  U51(x2, x4)
TIMES_IN(x1, x2, x3)  =  TIMES_IN(x1, x2)
U41(x1, x2, x3, x4)  =  U41(x4)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
CONVERT_IN(x1, x2, x3)  =  CONVERT_IN(x1, x2)
U21(x1, x2, x3, x4)  =  U21(x4)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1, x2)
U61(x1, x2, x3, x4)  =  U61(x4)
U11(x1, x2, x3, x4)  =  U11(x2, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 8 less nodes.

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

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

PLUS_IN(s(X), Y, s(Z)) → PLUS_IN(X, Y, Z)

The TRS R consists of the following rules:

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(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
              ↳ PiDP

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

PLUS_IN(s(X), Y, s(Z)) → PLUS_IN(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(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
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

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

PLUS_IN(s(X), Y) → PLUS_IN(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



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

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

TIMES_IN(s(X), Y, Z) → TIMES_IN(X, Y, U)

The TRS R consists of the following rules:

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)
TIMES_IN(x1, x2, x3)  =  TIMES_IN(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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

TIMES_IN(s(X), Y, Z) → TIMES_IN(X, Y, U)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
TIMES_IN(x1, x2, x3)  =  TIMES_IN(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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

TIMES_IN(s(X), Y) → TIMES_IN(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



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

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

CONVERT_IN(.(s(Y), XS), B, s(X)) → CONVERT_IN(.(Y, XS), B, X)
CONVERT_IN(.(0, XS), B, X) → CONVERT_IN(XS, B, Y)

The TRS R consists of the following rules:

convert_in(.(s(Y), XS), B, s(X)) → U3(Y, XS, B, X, convert_in(.(Y, XS), B, X))
convert_in(.(0, XS), B, X) → U1(XS, B, X, convert_in(XS, B, Y))
convert_in([], B, 0) → convert_out([], B, 0)
U1(XS, B, X, convert_out(XS, B, Y)) → U2(XS, B, X, times_in(Y, B, X))
times_in(s(X), Y, Z) → U5(X, Y, Z, times_in(X, Y, U))
times_in(0, Y, 0) → times_out(0, Y, 0)
U5(X, Y, Z, times_out(X, Y, U)) → U6(X, Y, Z, plus_in(Y, U, Z))
plus_in(s(X), Y, s(Z)) → U4(X, Y, Z, plus_in(X, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U4(X, Y, Z, plus_out(X, Y, Z)) → plus_out(s(X), Y, s(Z))
U6(X, Y, Z, plus_out(Y, U, Z)) → times_out(s(X), Y, Z)
U2(XS, B, X, times_out(Y, B, X)) → convert_out(.(0, XS), B, X)
U3(Y, XS, B, X, convert_out(.(Y, XS), B, X)) → convert_out(.(s(Y), XS), B, s(X))

The argument filtering Pi contains the following mapping:
convert_in(x1, x2, x3)  =  convert_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
0  =  0
U1(x1, x2, x3, x4)  =  U1(x2, x4)
[]  =  []
convert_out(x1, x2, x3)  =  convert_out(x3)
U2(x1, x2, x3, x4)  =  U2(x4)
times_in(x1, x2, x3)  =  times_in(x1, x2)
U5(x1, x2, x3, x4)  =  U5(x2, x4)
times_out(x1, x2, x3)  =  times_out(x3)
U6(x1, x2, x3, x4)  =  U6(x4)
plus_in(x1, x2, x3)  =  plus_in(x1, x2)
U4(x1, x2, x3, x4)  =  U4(x4)
plus_out(x1, x2, x3)  =  plus_out(x3)
CONVERT_IN(x1, x2, x3)  =  CONVERT_IN(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
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

CONVERT_IN(.(s(Y), XS), B, s(X)) → CONVERT_IN(.(Y, XS), B, X)
CONVERT_IN(.(0, XS), B, X) → CONVERT_IN(XS, B, Y)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
0  =  0
CONVERT_IN(x1, x2, x3)  =  CONVERT_IN(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
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof

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

CONVERT_IN(.(s(Y), XS), B) → CONVERT_IN(.(Y, XS), B)
CONVERT_IN(.(0, XS), B) → CONVERT_IN(XS, B)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

CONVERT_IN(.(s(Y), XS), B) → CONVERT_IN(.(Y, XS), B)
CONVERT_IN(.(0, XS), B) → CONVERT_IN(XS, B)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2·x1 + 2·x2   
POL(0) = 0   
POL(CONVERT_IN(x1, x2)) = 2·x1 + x2   
POL(s(x1)) = 2·x1   



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

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