Left Termination of the query pattern goal_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

tappend(nil, T, T).
tappend(node(nil, X, T2), T1, node(T1, X, T2)).
tappend(node(T1, X, nil), T2, node(T1, X, T2)).
tappend(node(T1, X, T2), T3, node(U, X, T2)) :- tappend(T1, T3, U).
tappend(node(T1, X, T2), T3, node(T1, X, U)) :- tappend(T2, T3, U).
s2t(s(X), node(T, Y, T)) :- s2t(X, T).
s2t(s(X), node(nil, Y, T)) :- s2t(X, T).
s2t(s(X), node(T, Y, nil)) :- s2t(X, T).
s2t(s(X), node(nil, Y, nil)).
s2t(0, nil).
goal(X) :- ','(s2t(X, T1), tappend(T1, T2, T3)).

Queries:

goal(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:

goal_in(X) → U6(X, s2t_in(X, T1))
s2t_in(0, nil) → s2t_out(0, nil)
s2t_in(s(X), node(nil, Y, nil)) → s2t_out(s(X), node(nil, Y, nil))
s2t_in(s(X), node(T, Y, nil)) → U5(X, T, Y, s2t_in(X, T))
s2t_in(s(X), node(nil, Y, T)) → U4(X, Y, T, s2t_in(X, T))
s2t_in(s(X), node(T, Y, T)) → U3(X, T, Y, s2t_in(X, T))
U3(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, T))
U4(X, Y, T, s2t_out(X, T)) → s2t_out(s(X), node(nil, Y, T))
U5(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, nil))
U6(X, s2t_out(X, T1)) → U7(X, tappend_in(T1, T2, T3))
tappend_in(node(T1, X, T2), T3, node(T1, X, U)) → U2(T1, X, T2, T3, U, tappend_in(T2, T3, U))
tappend_in(node(T1, X, T2), T3, node(U, X, T2)) → U1(T1, X, T2, T3, U, tappend_in(T1, T3, U))
tappend_in(node(T1, X, nil), T2, node(T1, X, T2)) → tappend_out(node(T1, X, nil), T2, node(T1, X, T2))
tappend_in(node(nil, X, T2), T1, node(T1, X, T2)) → tappend_out(node(nil, X, T2), T1, node(T1, X, T2))
tappend_in(nil, T, T) → tappend_out(nil, T, T)
U1(T1, X, T2, T3, U, tappend_out(T1, T3, U)) → tappend_out(node(T1, X, T2), T3, node(U, X, T2))
U2(T1, X, T2, T3, U, tappend_out(T2, T3, U)) → tappend_out(node(T1, X, T2), T3, node(T1, X, U))
U7(X, tappend_out(T1, T2, T3)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U6(x1, x2)  =  U6(x2)
s2t_in(x1, x2)  =  s2t_in(x1)
0  =  0
nil  =  nil
s2t_out(x1, x2)  =  s2t_out(x2)
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
U5(x1, x2, x3, x4)  =  U5(x4)
U4(x1, x2, x3, x4)  =  U4(x4)
U3(x1, x2, x3, x4)  =  U3(x4)
U7(x1, x2)  =  U7(x2)
tappend_in(x1, x2, x3)  =  tappend_in(x1)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
tappend_out(x1, x2, x3)  =  tappend_out
goal_out(x1)  =  goal_out

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:

goal_in(X) → U6(X, s2t_in(X, T1))
s2t_in(0, nil) → s2t_out(0, nil)
s2t_in(s(X), node(nil, Y, nil)) → s2t_out(s(X), node(nil, Y, nil))
s2t_in(s(X), node(T, Y, nil)) → U5(X, T, Y, s2t_in(X, T))
s2t_in(s(X), node(nil, Y, T)) → U4(X, Y, T, s2t_in(X, T))
s2t_in(s(X), node(T, Y, T)) → U3(X, T, Y, s2t_in(X, T))
U3(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, T))
U4(X, Y, T, s2t_out(X, T)) → s2t_out(s(X), node(nil, Y, T))
U5(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, nil))
U6(X, s2t_out(X, T1)) → U7(X, tappend_in(T1, T2, T3))
tappend_in(node(T1, X, T2), T3, node(T1, X, U)) → U2(T1, X, T2, T3, U, tappend_in(T2, T3, U))
tappend_in(node(T1, X, T2), T3, node(U, X, T2)) → U1(T1, X, T2, T3, U, tappend_in(T1, T3, U))
tappend_in(node(T1, X, nil), T2, node(T1, X, T2)) → tappend_out(node(T1, X, nil), T2, node(T1, X, T2))
tappend_in(node(nil, X, T2), T1, node(T1, X, T2)) → tappend_out(node(nil, X, T2), T1, node(T1, X, T2))
tappend_in(nil, T, T) → tappend_out(nil, T, T)
U1(T1, X, T2, T3, U, tappend_out(T1, T3, U)) → tappend_out(node(T1, X, T2), T3, node(U, X, T2))
U2(T1, X, T2, T3, U, tappend_out(T2, T3, U)) → tappend_out(node(T1, X, T2), T3, node(T1, X, U))
U7(X, tappend_out(T1, T2, T3)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U6(x1, x2)  =  U6(x2)
s2t_in(x1, x2)  =  s2t_in(x1)
0  =  0
nil  =  nil
s2t_out(x1, x2)  =  s2t_out(x2)
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
U5(x1, x2, x3, x4)  =  U5(x4)
U4(x1, x2, x3, x4)  =  U4(x4)
U3(x1, x2, x3, x4)  =  U3(x4)
U7(x1, x2)  =  U7(x2)
tappend_in(x1, x2, x3)  =  tappend_in(x1)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
tappend_out(x1, x2, x3)  =  tappend_out
goal_out(x1)  =  goal_out


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

GOAL_IN(X) → U61(X, s2t_in(X, T1))
GOAL_IN(X) → S2T_IN(X, T1)
S2T_IN(s(X), node(T, Y, nil)) → U51(X, T, Y, s2t_in(X, T))
S2T_IN(s(X), node(T, Y, nil)) → S2T_IN(X, T)
S2T_IN(s(X), node(nil, Y, T)) → U41(X, Y, T, s2t_in(X, T))
S2T_IN(s(X), node(nil, Y, T)) → S2T_IN(X, T)
S2T_IN(s(X), node(T, Y, T)) → U31(X, T, Y, s2t_in(X, T))
S2T_IN(s(X), node(T, Y, T)) → S2T_IN(X, T)
U61(X, s2t_out(X, T1)) → U71(X, tappend_in(T1, T2, T3))
U61(X, s2t_out(X, T1)) → TAPPEND_IN(T1, T2, T3)
TAPPEND_IN(node(T1, X, T2), T3, node(T1, X, U)) → U21(T1, X, T2, T3, U, tappend_in(T2, T3, U))
TAPPEND_IN(node(T1, X, T2), T3, node(T1, X, U)) → TAPPEND_IN(T2, T3, U)
TAPPEND_IN(node(T1, X, T2), T3, node(U, X, T2)) → U11(T1, X, T2, T3, U, tappend_in(T1, T3, U))
TAPPEND_IN(node(T1, X, T2), T3, node(U, X, T2)) → TAPPEND_IN(T1, T3, U)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T1))
s2t_in(0, nil) → s2t_out(0, nil)
s2t_in(s(X), node(nil, Y, nil)) → s2t_out(s(X), node(nil, Y, nil))
s2t_in(s(X), node(T, Y, nil)) → U5(X, T, Y, s2t_in(X, T))
s2t_in(s(X), node(nil, Y, T)) → U4(X, Y, T, s2t_in(X, T))
s2t_in(s(X), node(T, Y, T)) → U3(X, T, Y, s2t_in(X, T))
U3(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, T))
U4(X, Y, T, s2t_out(X, T)) → s2t_out(s(X), node(nil, Y, T))
U5(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, nil))
U6(X, s2t_out(X, T1)) → U7(X, tappend_in(T1, T2, T3))
tappend_in(node(T1, X, T2), T3, node(T1, X, U)) → U2(T1, X, T2, T3, U, tappend_in(T2, T3, U))
tappend_in(node(T1, X, T2), T3, node(U, X, T2)) → U1(T1, X, T2, T3, U, tappend_in(T1, T3, U))
tappend_in(node(T1, X, nil), T2, node(T1, X, T2)) → tappend_out(node(T1, X, nil), T2, node(T1, X, T2))
tappend_in(node(nil, X, T2), T1, node(T1, X, T2)) → tappend_out(node(nil, X, T2), T1, node(T1, X, T2))
tappend_in(nil, T, T) → tappend_out(nil, T, T)
U1(T1, X, T2, T3, U, tappend_out(T1, T3, U)) → tappend_out(node(T1, X, T2), T3, node(U, X, T2))
U2(T1, X, T2, T3, U, tappend_out(T2, T3, U)) → tappend_out(node(T1, X, T2), T3, node(T1, X, U))
U7(X, tappend_out(T1, T2, T3)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U6(x1, x2)  =  U6(x2)
s2t_in(x1, x2)  =  s2t_in(x1)
0  =  0
nil  =  nil
s2t_out(x1, x2)  =  s2t_out(x2)
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
U5(x1, x2, x3, x4)  =  U5(x4)
U4(x1, x2, x3, x4)  =  U4(x4)
U3(x1, x2, x3, x4)  =  U3(x4)
U7(x1, x2)  =  U7(x2)
tappend_in(x1, x2, x3)  =  tappend_in(x1)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
tappend_out(x1, x2, x3)  =  tappend_out
goal_out(x1)  =  goal_out
U71(x1, x2)  =  U71(x2)
U51(x1, x2, x3, x4)  =  U51(x4)
U61(x1, x2)  =  U61(x2)
U31(x1, x2, x3, x4)  =  U31(x4)
TAPPEND_IN(x1, x2, x3)  =  TAPPEND_IN(x1)
U41(x1, x2, x3, x4)  =  U41(x4)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x6)
S2T_IN(x1, x2)  =  S2T_IN(x1)
GOAL_IN(x1)  =  GOAL_IN(x1)
U11(x1, x2, x3, x4, x5, x6)  =  U11(x6)

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:

GOAL_IN(X) → U61(X, s2t_in(X, T1))
GOAL_IN(X) → S2T_IN(X, T1)
S2T_IN(s(X), node(T, Y, nil)) → U51(X, T, Y, s2t_in(X, T))
S2T_IN(s(X), node(T, Y, nil)) → S2T_IN(X, T)
S2T_IN(s(X), node(nil, Y, T)) → U41(X, Y, T, s2t_in(X, T))
S2T_IN(s(X), node(nil, Y, T)) → S2T_IN(X, T)
S2T_IN(s(X), node(T, Y, T)) → U31(X, T, Y, s2t_in(X, T))
S2T_IN(s(X), node(T, Y, T)) → S2T_IN(X, T)
U61(X, s2t_out(X, T1)) → U71(X, tappend_in(T1, T2, T3))
U61(X, s2t_out(X, T1)) → TAPPEND_IN(T1, T2, T3)
TAPPEND_IN(node(T1, X, T2), T3, node(T1, X, U)) → U21(T1, X, T2, T3, U, tappend_in(T2, T3, U))
TAPPEND_IN(node(T1, X, T2), T3, node(T1, X, U)) → TAPPEND_IN(T2, T3, U)
TAPPEND_IN(node(T1, X, T2), T3, node(U, X, T2)) → U11(T1, X, T2, T3, U, tappend_in(T1, T3, U))
TAPPEND_IN(node(T1, X, T2), T3, node(U, X, T2)) → TAPPEND_IN(T1, T3, U)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T1))
s2t_in(0, nil) → s2t_out(0, nil)
s2t_in(s(X), node(nil, Y, nil)) → s2t_out(s(X), node(nil, Y, nil))
s2t_in(s(X), node(T, Y, nil)) → U5(X, T, Y, s2t_in(X, T))
s2t_in(s(X), node(nil, Y, T)) → U4(X, Y, T, s2t_in(X, T))
s2t_in(s(X), node(T, Y, T)) → U3(X, T, Y, s2t_in(X, T))
U3(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, T))
U4(X, Y, T, s2t_out(X, T)) → s2t_out(s(X), node(nil, Y, T))
U5(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, nil))
U6(X, s2t_out(X, T1)) → U7(X, tappend_in(T1, T2, T3))
tappend_in(node(T1, X, T2), T3, node(T1, X, U)) → U2(T1, X, T2, T3, U, tappend_in(T2, T3, U))
tappend_in(node(T1, X, T2), T3, node(U, X, T2)) → U1(T1, X, T2, T3, U, tappend_in(T1, T3, U))
tappend_in(node(T1, X, nil), T2, node(T1, X, T2)) → tappend_out(node(T1, X, nil), T2, node(T1, X, T2))
tappend_in(node(nil, X, T2), T1, node(T1, X, T2)) → tappend_out(node(nil, X, T2), T1, node(T1, X, T2))
tappend_in(nil, T, T) → tappend_out(nil, T, T)
U1(T1, X, T2, T3, U, tappend_out(T1, T3, U)) → tappend_out(node(T1, X, T2), T3, node(U, X, T2))
U2(T1, X, T2, T3, U, tappend_out(T2, T3, U)) → tappend_out(node(T1, X, T2), T3, node(T1, X, U))
U7(X, tappend_out(T1, T2, T3)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U6(x1, x2)  =  U6(x2)
s2t_in(x1, x2)  =  s2t_in(x1)
0  =  0
nil  =  nil
s2t_out(x1, x2)  =  s2t_out(x2)
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
U5(x1, x2, x3, x4)  =  U5(x4)
U4(x1, x2, x3, x4)  =  U4(x4)
U3(x1, x2, x3, x4)  =  U3(x4)
U7(x1, x2)  =  U7(x2)
tappend_in(x1, x2, x3)  =  tappend_in(x1)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
tappend_out(x1, x2, x3)  =  tappend_out
goal_out(x1)  =  goal_out
U71(x1, x2)  =  U71(x2)
U51(x1, x2, x3, x4)  =  U51(x4)
U61(x1, x2)  =  U61(x2)
U31(x1, x2, x3, x4)  =  U31(x4)
TAPPEND_IN(x1, x2, x3)  =  TAPPEND_IN(x1)
U41(x1, x2, x3, x4)  =  U41(x4)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x6)
S2T_IN(x1, x2)  =  S2T_IN(x1)
GOAL_IN(x1)  =  GOAL_IN(x1)
U11(x1, x2, x3, x4, x5, x6)  =  U11(x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 9 less nodes.

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

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

TAPPEND_IN(node(T1, X, T2), T3, node(U, X, T2)) → TAPPEND_IN(T1, T3, U)
TAPPEND_IN(node(T1, X, T2), T3, node(T1, X, U)) → TAPPEND_IN(T2, T3, U)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T1))
s2t_in(0, nil) → s2t_out(0, nil)
s2t_in(s(X), node(nil, Y, nil)) → s2t_out(s(X), node(nil, Y, nil))
s2t_in(s(X), node(T, Y, nil)) → U5(X, T, Y, s2t_in(X, T))
s2t_in(s(X), node(nil, Y, T)) → U4(X, Y, T, s2t_in(X, T))
s2t_in(s(X), node(T, Y, T)) → U3(X, T, Y, s2t_in(X, T))
U3(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, T))
U4(X, Y, T, s2t_out(X, T)) → s2t_out(s(X), node(nil, Y, T))
U5(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, nil))
U6(X, s2t_out(X, T1)) → U7(X, tappend_in(T1, T2, T3))
tappend_in(node(T1, X, T2), T3, node(T1, X, U)) → U2(T1, X, T2, T3, U, tappend_in(T2, T3, U))
tappend_in(node(T1, X, T2), T3, node(U, X, T2)) → U1(T1, X, T2, T3, U, tappend_in(T1, T3, U))
tappend_in(node(T1, X, nil), T2, node(T1, X, T2)) → tappend_out(node(T1, X, nil), T2, node(T1, X, T2))
tappend_in(node(nil, X, T2), T1, node(T1, X, T2)) → tappend_out(node(nil, X, T2), T1, node(T1, X, T2))
tappend_in(nil, T, T) → tappend_out(nil, T, T)
U1(T1, X, T2, T3, U, tappend_out(T1, T3, U)) → tappend_out(node(T1, X, T2), T3, node(U, X, T2))
U2(T1, X, T2, T3, U, tappend_out(T2, T3, U)) → tappend_out(node(T1, X, T2), T3, node(T1, X, U))
U7(X, tappend_out(T1, T2, T3)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U6(x1, x2)  =  U6(x2)
s2t_in(x1, x2)  =  s2t_in(x1)
0  =  0
nil  =  nil
s2t_out(x1, x2)  =  s2t_out(x2)
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
U5(x1, x2, x3, x4)  =  U5(x4)
U4(x1, x2, x3, x4)  =  U4(x4)
U3(x1, x2, x3, x4)  =  U3(x4)
U7(x1, x2)  =  U7(x2)
tappend_in(x1, x2, x3)  =  tappend_in(x1)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
tappend_out(x1, x2, x3)  =  tappend_out
goal_out(x1)  =  goal_out
TAPPEND_IN(x1, x2, x3)  =  TAPPEND_IN(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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

TAPPEND_IN(node(T1, X, T2), T3, node(U, X, T2)) → TAPPEND_IN(T1, T3, U)
TAPPEND_IN(node(T1, X, T2), T3, node(T1, X, U)) → TAPPEND_IN(T2, T3, U)

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

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

TAPPEND_IN(node(T1, T2)) → TAPPEND_IN(T1)
TAPPEND_IN(node(T1, T2)) → TAPPEND_IN(T2)

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

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

S2T_IN(s(X), node(T, Y, T)) → S2T_IN(X, T)
S2T_IN(s(X), node(T, Y, nil)) → S2T_IN(X, T)
S2T_IN(s(X), node(nil, Y, T)) → S2T_IN(X, T)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T1))
s2t_in(0, nil) → s2t_out(0, nil)
s2t_in(s(X), node(nil, Y, nil)) → s2t_out(s(X), node(nil, Y, nil))
s2t_in(s(X), node(T, Y, nil)) → U5(X, T, Y, s2t_in(X, T))
s2t_in(s(X), node(nil, Y, T)) → U4(X, Y, T, s2t_in(X, T))
s2t_in(s(X), node(T, Y, T)) → U3(X, T, Y, s2t_in(X, T))
U3(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, T))
U4(X, Y, T, s2t_out(X, T)) → s2t_out(s(X), node(nil, Y, T))
U5(X, T, Y, s2t_out(X, T)) → s2t_out(s(X), node(T, Y, nil))
U6(X, s2t_out(X, T1)) → U7(X, tappend_in(T1, T2, T3))
tappend_in(node(T1, X, T2), T3, node(T1, X, U)) → U2(T1, X, T2, T3, U, tappend_in(T2, T3, U))
tappend_in(node(T1, X, T2), T3, node(U, X, T2)) → U1(T1, X, T2, T3, U, tappend_in(T1, T3, U))
tappend_in(node(T1, X, nil), T2, node(T1, X, T2)) → tappend_out(node(T1, X, nil), T2, node(T1, X, T2))
tappend_in(node(nil, X, T2), T1, node(T1, X, T2)) → tappend_out(node(nil, X, T2), T1, node(T1, X, T2))
tappend_in(nil, T, T) → tappend_out(nil, T, T)
U1(T1, X, T2, T3, U, tappend_out(T1, T3, U)) → tappend_out(node(T1, X, T2), T3, node(U, X, T2))
U2(T1, X, T2, T3, U, tappend_out(T2, T3, U)) → tappend_out(node(T1, X, T2), T3, node(T1, X, U))
U7(X, tappend_out(T1, T2, T3)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U6(x1, x2)  =  U6(x2)
s2t_in(x1, x2)  =  s2t_in(x1)
0  =  0
nil  =  nil
s2t_out(x1, x2)  =  s2t_out(x2)
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
U5(x1, x2, x3, x4)  =  U5(x4)
U4(x1, x2, x3, x4)  =  U4(x4)
U3(x1, x2, x3, x4)  =  U3(x4)
U7(x1, x2)  =  U7(x2)
tappend_in(x1, x2, x3)  =  tappend_in(x1)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
tappend_out(x1, x2, x3)  =  tappend_out
goal_out(x1)  =  goal_out
S2T_IN(x1, x2)  =  S2T_IN(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

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

S2T_IN(s(X), node(T, Y, T)) → S2T_IN(X, T)
S2T_IN(s(X), node(T, Y, nil)) → S2T_IN(X, T)
S2T_IN(s(X), node(nil, Y, T)) → S2T_IN(X, T)

R is empty.
The argument filtering Pi contains the following mapping:
nil  =  nil
s(x1)  =  s(x1)
node(x1, x2, x3)  =  node(x1, x3)
S2T_IN(x1, x2)  =  S2T_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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof

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

S2T_IN(s(X)) → S2T_IN(X)

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: