Left Termination of the query pattern som3_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

som3([], Bs, Bs).
som3(As, [], As).
som3(.(A, As), .(B, Bs), .(+(A, B), Cs)) :- som3(As, Bs, Cs).
som4_1(As, Bs, Cs, Ds) :- ','(som3(As, Bs, Es), som3(Es, Cs, Ds)).
som4_2(As, Bs, Cs, Ds) :- ','(som3(Es, Cs, Ds), som3(As, Bs, Es)).

Queries:

som3(g,a,a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
som3_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)

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:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)


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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(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:

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_GAA(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

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

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

The TRS R consists of the following rules:

som3_in_gaa([], Bs, Bs) → som3_out_gaa([], Bs, Bs)
som3_in_gaa(As, [], As) → som3_out_gaa(As, [], As)
som3_in_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs)) → U1_gaa(A, As, B, Bs, Cs, som3_in_gaa(As, Bs, Cs))
U1_gaa(A, As, B, Bs, Cs, som3_out_gaa(As, Bs, Cs)) → som3_out_gaa(.(A, As), .(B, Bs), .(+(A, B), Cs))

The argument filtering Pi contains the following mapping:
som3_in_gaa(x1, x2, x3)  =  som3_in_gaa(x1)
[]  =  []
som3_out_gaa(x1, x2, x3)  =  som3_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

SOM3_IN_GAA(.(A, As), .(B, Bs), .(+(A, B), Cs)) → SOM3_IN_GAA(As, Bs, Cs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
SOM3_IN_GAA(x1, x2, x3)  =  SOM3_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

SOM3_IN_GAA(.(A, As)) → SOM3_IN_GAA(As)

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: