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



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

tree(nil).
tree(node(L, X, R)) :- ','(tree(L), tree(R)).
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, T), tree(T)).

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, T))
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, T)) → U7(X, tree_in(T))
tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))
U7(X, tree_out(T)) → 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)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
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, T))
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, T)) → U7(X, tree_in(T))
tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))
U7(X, tree_out(T)) → 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)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
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, T))
GOAL_IN(X) → S2T_IN(X, T)
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, T)) → U71(X, tree_in(T))
U61(X, s2t_out(X, T)) → TREE_IN(T)
TREE_IN(node(L, X, R)) → U11(L, X, R, tree_in(L))
TREE_IN(node(L, X, R)) → TREE_IN(L)
U11(L, X, R, tree_out(L)) → U21(L, X, R, tree_in(R))
U11(L, X, R, tree_out(L)) → TREE_IN(R)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T))
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, T)) → U7(X, tree_in(T))
tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))
U7(X, tree_out(T)) → 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)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
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)
TREE_IN(x1)  =  TREE_IN(x1)
U41(x1, x2, x3, x4)  =  U41(x4)
U21(x1, x2, x3, x4)  =  U21(x4)
S2T_IN(x1, x2)  =  S2T_IN(x1)
GOAL_IN(x1)  =  GOAL_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x3, 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:

GOAL_IN(X) → U61(X, s2t_in(X, T))
GOAL_IN(X) → S2T_IN(X, T)
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, T)) → U71(X, tree_in(T))
U61(X, s2t_out(X, T)) → TREE_IN(T)
TREE_IN(node(L, X, R)) → U11(L, X, R, tree_in(L))
TREE_IN(node(L, X, R)) → TREE_IN(L)
U11(L, X, R, tree_out(L)) → U21(L, X, R, tree_in(R))
U11(L, X, R, tree_out(L)) → TREE_IN(R)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T))
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, T)) → U7(X, tree_in(T))
tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))
U7(X, tree_out(T)) → 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)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
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)
TREE_IN(x1)  =  TREE_IN(x1)
U41(x1, x2, x3, x4)  =  U41(x4)
U21(x1, x2, x3, x4)  =  U21(x4)
S2T_IN(x1, x2)  =  S2T_IN(x1)
GOAL_IN(x1)  =  GOAL_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 8 less nodes.

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

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

TREE_IN(node(L, X, R)) → U11(L, X, R, tree_in(L))
U11(L, X, R, tree_out(L)) → TREE_IN(R)
TREE_IN(node(L, X, R)) → TREE_IN(L)

The TRS R consists of the following rules:

goal_in(X) → U6(X, s2t_in(X, T))
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, T)) → U7(X, tree_in(T))
tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))
U7(X, tree_out(T)) → 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)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
goal_out(x1)  =  goal_out
TREE_IN(x1)  =  TREE_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x3, x4)

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:

TREE_IN(node(L, X, R)) → U11(L, X, R, tree_in(L))
U11(L, X, R, tree_out(L)) → TREE_IN(R)
TREE_IN(node(L, X, R)) → TREE_IN(L)

The TRS R consists of the following rules:

tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))

The argument filtering Pi contains the following mapping:
nil  =  nil
node(x1, x2, x3)  =  node(x1, x3)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
TREE_IN(x1)  =  TREE_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x3, x4)

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:

U11(R, tree_out) → TREE_IN(R)
TREE_IN(node(L, R)) → TREE_IN(L)
TREE_IN(node(L, R)) → U11(R, tree_in(L))

The TRS R consists of the following rules:

tree_in(node(L, R)) → U1(R, tree_in(L))
tree_in(nil) → tree_out
U1(R, tree_out) → U2(tree_in(R))
U2(tree_out) → tree_out

The set Q consists of the following terms:

tree_in(x0)
U1(x0, x1)
U2(x0)

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, T))
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, T)) → U7(X, tree_in(T))
tree_in(node(L, X, R)) → U1(L, X, R, tree_in(L))
tree_in(nil) → tree_out(nil)
U1(L, X, R, tree_out(L)) → U2(L, X, R, tree_in(R))
U2(L, X, R, tree_out(R)) → tree_out(node(L, X, R))
U7(X, tree_out(T)) → 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)
tree_in(x1)  =  tree_in(x1)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
tree_out(x1)  =  tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
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: