Left Termination of the query pattern som3(b,f,f) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof

som33({}0, Bs, Bs).
som33(As, {}0, As).
som33(.2(A, As), .2(B, Bs), .2(A + B, Cs)) :- som33(As, Bs, Cs).
som414(As, Bs, Cs, Ds) :- som33(As, Bs, Es), som33(Es, Cs, Ds).
som424(As, Bs, Cs, Ds) :- som33(Es, Cs, Ds), som33(As, Bs, Es).


The clauses

som414(As, Bs, Cs, Ds) :- som33(As, Bs, Es), som33(Es, Cs, Ds).
som424(As, Bs, Cs, Ds) :- som33(Es, Cs, Ds), som33(As, Bs, Es).

can be ignored, as they are not needed by any of the given querys.

Deleting these clauses results in the following prolog program:

som33({}0, Bs, Bs).
som33(As, {}0, As).
som33(.2(A, As), .2(B, Bs), .2(A + B, Cs)) :- som33(As, Bs, Cs).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof

som33({}0, Bs, Bs).
som33(As, {}0, As).
som33(.2(A, As), .2(B, Bs), .2(A + B, Cs)) :- som33(As, Bs, Cs).


With regard to the inferred argument filtering the predicates were used in the following modes:
som33: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


som3_3_in_gaa3([]_0, Bs, Bs) -> som3_3_out_gaa3([]_0, Bs, Bs)
som3_3_in_gaa3(As, []_0, As) -> som3_3_out_gaa3(As, []_0, As)
som3_3_in_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_out_gaa3(As, Bs, Cs)) -> som3_3_out_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_3_in_gaa3(x1, x2, x3)  =  som3_3_in_gaa1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
som3_3_out_gaa3(x1, x2, x3)  =  som3_3_out_gaa
if_som3_3_in_1_gaa6(x1, x2, x3, x4, x5, x6)  =  if_som3_3_in_1_gaa1(x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof

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

som3_3_in_gaa3([]_0, Bs, Bs) -> som3_3_out_gaa3([]_0, Bs, Bs)
som3_3_in_gaa3(As, []_0, As) -> som3_3_out_gaa3(As, []_0, As)
som3_3_in_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_out_gaa3(As, Bs, Cs)) -> som3_3_out_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_3_in_gaa3(x1, x2, x3)  =  som3_3_in_gaa1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
som3_3_out_gaa3(x1, x2, x3)  =  som3_3_out_gaa
if_som3_3_in_1_gaa6(x1, x2, x3, x4, x5, x6)  =  if_som3_3_in_1_gaa1(x6)


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

SOM3_3_IN_GAA3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> IF_SOM3_3_IN_1_GAA6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
SOM3_3_IN_GAA3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> SOM3_3_IN_GAA3(As, Bs, Cs)

The TRS R consists of the following rules:

som3_3_in_gaa3([]_0, Bs, Bs) -> som3_3_out_gaa3([]_0, Bs, Bs)
som3_3_in_gaa3(As, []_0, As) -> som3_3_out_gaa3(As, []_0, As)
som3_3_in_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_out_gaa3(As, Bs, Cs)) -> som3_3_out_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_3_in_gaa3(x1, x2, x3)  =  som3_3_in_gaa1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
som3_3_out_gaa3(x1, x2, x3)  =  som3_3_out_gaa
if_som3_3_in_1_gaa6(x1, x2, x3, x4, x5, x6)  =  if_som3_3_in_1_gaa1(x6)
IF_SOM3_3_IN_1_GAA6(x1, x2, x3, x4, x5, x6)  =  IF_SOM3_3_IN_1_GAA1(x6)
SOM3_3_IN_GAA3(x1, x2, x3)  =  SOM3_3_IN_GAA1(x1)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof

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

SOM3_3_IN_GAA3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> IF_SOM3_3_IN_1_GAA6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
SOM3_3_IN_GAA3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> SOM3_3_IN_GAA3(As, Bs, Cs)

The TRS R consists of the following rules:

som3_3_in_gaa3([]_0, Bs, Bs) -> som3_3_out_gaa3([]_0, Bs, Bs)
som3_3_in_gaa3(As, []_0, As) -> som3_3_out_gaa3(As, []_0, As)
som3_3_in_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_out_gaa3(As, Bs, Cs)) -> som3_3_out_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_3_in_gaa3(x1, x2, x3)  =  som3_3_in_gaa1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
som3_3_out_gaa3(x1, x2, x3)  =  som3_3_out_gaa
if_som3_3_in_1_gaa6(x1, x2, x3, x4, x5, x6)  =  if_som3_3_in_1_gaa1(x6)
IF_SOM3_3_IN_1_GAA6(x1, x2, x3, x4, x5, x6)  =  IF_SOM3_3_IN_1_GAA1(x6)
SOM3_3_IN_GAA3(x1, x2, x3)  =  SOM3_3_IN_GAA1(x1)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof

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

SOM3_3_IN_GAA3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> SOM3_3_IN_GAA3(As, Bs, Cs)

The TRS R consists of the following rules:

som3_3_in_gaa3([]_0, Bs, Bs) -> som3_3_out_gaa3([]_0, Bs, Bs)
som3_3_in_gaa3(As, []_0, As) -> som3_3_out_gaa3(As, []_0, As)
som3_3_in_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_in_gaa3(As, Bs, Cs))
if_som3_3_in_1_gaa6(A, As, B, Bs, Cs, som3_3_out_gaa3(As, Bs, Cs)) -> som3_3_out_gaa3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_3_in_gaa3(x1, x2, x3)  =  som3_3_in_gaa1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
som3_3_out_gaa3(x1, x2, x3)  =  som3_3_out_gaa
if_som3_3_in_1_gaa6(x1, x2, x3, x4, x5, x6)  =  if_som3_3_in_1_gaa1(x6)
SOM3_3_IN_GAA3(x1, x2, x3)  =  SOM3_3_IN_GAA1(x1)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof

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

SOM3_3_IN_GAA3(._22(A, As), ._22(B, Bs), ._22(+2(A, B), Cs)) -> SOM3_3_IN_GAA3(As, Bs, Cs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
SOM3_3_IN_GAA3(x1, x2, x3)  =  SOM3_3_IN_GAA1(x1)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof
                        ↳ QDP
                          ↳ QDPSizeChangeProof

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

SOM3_3_IN_GAA1(._22(A, As)) -> SOM3_3_IN_GAA1(As)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SOM3_3_IN_GAA1}.
By using the subterm criterion together with the size-change analysis 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: