Left Termination of the query pattern sum_in_3(g, a, g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
  ↳19 PrologToPiTRSProof (⇐)
    ↳20 PiTRS
    ↳21 DependencyPairsProof (⇔)
    ↳22 PiDP
    ↳23 DependencyGraphProof (⇔)
    ↳24 AND
      ↳25 PiDP
        ↳26 UsableRulesProof (⇔)
        ↳27 PiDP
        ↳28 PiDPToQDPProof (⇐)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔)
        ↳31 TRUE
      ↳32 PiDP
        ↳33 UsableRulesProof (⇔)
        ↳34 PiDP
        ↳35 PiDPToQDPProof (⇐)
        ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔)
        ↳38 TRUE

(0) Obligation:

Clauses:

sum([], [], []).
sum(.(X1, Y1), .(X2, Y2), .(X3, Y3)) :- ','(add(X1, X2, X3), sum(Y1, Y2, Y3)).
add(0, X, X).
add(s(X), Y, s(Z)) :- add(X, Y, Z).

Queries:

sum(g,a,g).

(1) PrologToPiTRSProof (SOUND transformation)

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

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x1, x2, x5, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x1, x2, x3, x5, x6, x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x1, x2, x5, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x1, x2, x3, x5, x6, x7)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → ADD_IN_GAG(X1, X2, X3)
ADD_IN_GAG(s(X), Y, s(Z)) → U3_GAG(X, Y, Z, add_in_gag(X, Y, Z))
ADD_IN_GAG(s(X), Y, s(Z)) → ADD_IN_GAG(X, Y, Z)
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_GAG(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x1, x2, x5, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x1, x2, x3, x5, x6, x7)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x1, x2, x5, x6, x7)
ADD_IN_GAG(x1, x2, x3)  =  ADD_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4)  =  U3_GAG(x1, x3, x4)
U2_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAG(x1, x2, x3, x5, x6, x7)

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

(4) Obligation:

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

SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → ADD_IN_GAG(X1, X2, X3)
ADD_IN_GAG(s(X), Y, s(Z)) → U3_GAG(X, Y, Z, add_in_gag(X, Y, Z))
ADD_IN_GAG(s(X), Y, s(Z)) → ADD_IN_GAG(X, Y, Z)
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_GAG(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x1, x2, x5, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x1, x2, x3, x5, x6, x7)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x1, x2, x5, x6, x7)
ADD_IN_GAG(x1, x2, x3)  =  ADD_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4)  =  U3_GAG(x1, x3, x4)
U2_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAG(x1, x2, x3, x5, x6, x7)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x1, x2, x5, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x1, x2, x3, x5, x6, x7)
ADD_IN_GAG(x1, x2, x3)  =  ADD_IN_GAG(x1, x3)

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

(8) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(9) Obligation:

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

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

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

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

(10) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(11) Obligation:

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

ADD_IN_GAG(s(X), s(Z)) → ADD_IN_GAG(X, Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • ADD_IN_GAG(s(X), s(Z)) → ADD_IN_GAG(X, Z)
    The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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

U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x1, x2, x5, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x1, x2, x3, x5, x6, x7)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x1, x2, x5, x6, x7)

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

(15) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(16) Obligation:

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

U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))

The TRS R consists of the following rules:

add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x1, x2, x3)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x1, x3, x4)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x1, x2, x5, x6, x7)

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

(17) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(18) Obligation:

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

U1_GAG(X1, Y1, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y3)
SUM_IN_GAG(.(X1, Y1), .(X3, Y3)) → U1_GAG(X1, Y1, X3, Y3, add_in_gag(X1, X3))

The TRS R consists of the following rules:

add_in_gag(0, X) → add_out_gag(0, X, X)
add_in_gag(s(X), s(Z)) → U3_gag(X, Z, add_in_gag(X, Z))
U3_gag(X, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))

The set Q consists of the following terms:

add_in_gag(x0, x1)
U3_gag(x0, x1, x2)

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

(19) PrologToPiTRSProof (SOUND transformation)

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

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x2, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x3, x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(20) Obligation:

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

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x2, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x3, x7)

(21) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → ADD_IN_GAG(X1, X2, X3)
ADD_IN_GAG(s(X), Y, s(Z)) → U3_GAG(X, Y, Z, add_in_gag(X, Y, Z))
ADD_IN_GAG(s(X), Y, s(Z)) → ADD_IN_GAG(X, Y, Z)
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_GAG(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x2, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x3, x7)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x2, x6, x7)
ADD_IN_GAG(x1, x2, x3)  =  ADD_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4)  =  U3_GAG(x4)
U2_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAG(x3, x7)

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

(22) Obligation:

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

SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → ADD_IN_GAG(X1, X2, X3)
ADD_IN_GAG(s(X), Y, s(Z)) → U3_GAG(X, Y, Z, add_in_gag(X, Y, Z))
ADD_IN_GAG(s(X), Y, s(Z)) → ADD_IN_GAG(X, Y, Z)
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_GAG(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x2, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x3, x7)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x2, x6, x7)
ADD_IN_GAG(x1, x2, x3)  =  ADD_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4)  =  U3_GAG(x4)
U2_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAG(x3, x7)

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(24) Complex Obligation (AND)

(25) Obligation:

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

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

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x2, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x3, x7)
ADD_IN_GAG(x1, x2, x3)  =  ADD_IN_GAG(x1, x3)

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

(26) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(27) Obligation:

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

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

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

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

(28) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(29) Obligation:

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

ADD_IN_GAG(s(X), s(Z)) → ADD_IN_GAG(X, Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(30) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • ADD_IN_GAG(s(X), s(Z)) → ADD_IN_GAG(X, Z)
    The graph contains the following edges 1 > 1, 2 > 2

(31) TRUE

(32) Obligation:

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

U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))

The TRS R consists of the following rules:

sum_in_gag([], [], []) → sum_out_gag([], [], [])
sum_in_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_gag(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))
add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))
U1_gag(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_in_gag(Y1, Y2, Y3))
U2_gag(X1, Y1, X2, Y2, X3, Y3, sum_out_gag(Y1, Y2, Y3)) → sum_out_gag(.(X1, Y1), .(X2, Y2), .(X3, Y3))

The argument filtering Pi contains the following mapping:
sum_in_gag(x1, x2, x3)  =  sum_in_gag(x1, x3)
[]  =  []
sum_out_gag(x1, x2, x3)  =  sum_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6, x7)  =  U1_gag(x2, x6, x7)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
U2_gag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gag(x3, x7)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x2, x6, x7)

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

(33) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(34) Obligation:

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

U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_out_gag(X1, X2, X3)) → SUM_IN_GAG(Y1, Y2, Y3)
SUM_IN_GAG(.(X1, Y1), .(X2, Y2), .(X3, Y3)) → U1_GAG(X1, Y1, X2, Y2, X3, Y3, add_in_gag(X1, X2, X3))

The TRS R consists of the following rules:

add_in_gag(0, X, X) → add_out_gag(0, X, X)
add_in_gag(s(X), Y, s(Z)) → U3_gag(X, Y, Z, add_in_gag(X, Y, Z))
U3_gag(X, Y, Z, add_out_gag(X, Y, Z)) → add_out_gag(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
add_in_gag(x1, x2, x3)  =  add_in_gag(x1, x3)
0  =  0
add_out_gag(x1, x2, x3)  =  add_out_gag(x2)
s(x1)  =  s(x1)
U3_gag(x1, x2, x3, x4)  =  U3_gag(x4)
SUM_IN_GAG(x1, x2, x3)  =  SUM_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U1_GAG(x2, x6, x7)

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

(35) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(36) Obligation:

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

U1_GAG(Y1, Y3, add_out_gag(X2)) → SUM_IN_GAG(Y1, Y3)
SUM_IN_GAG(.(X1, Y1), .(X3, Y3)) → U1_GAG(Y1, Y3, add_in_gag(X1, X3))

The TRS R consists of the following rules:

add_in_gag(0, X) → add_out_gag(X)
add_in_gag(s(X), s(Z)) → U3_gag(add_in_gag(X, Z))
U3_gag(add_out_gag(Y)) → add_out_gag(Y)

The set Q consists of the following terms:

add_in_gag(x0, x1)
U3_gag(x0)

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

(37) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • SUM_IN_GAG(.(X1, Y1), .(X3, Y3)) → U1_GAG(Y1, Y3, add_in_gag(X1, X3))
    The graph contains the following edges 1 > 1, 2 > 2

  • U1_GAG(Y1, Y3, add_out_gag(X2)) → SUM_IN_GAG(Y1, Y3)
    The graph contains the following edges 1 >= 1, 2 >= 2

(38) TRUE