(0) Obligation:

Clauses:

p(X, Y).
map(.(X, Xs), .(Y, Ys)) :- ','(p(X, Y), map(Xs, Ys)).
map([], []).

Queries:

map(g,a).

(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:
map_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa

(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:

MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → P_IN_AA(X, Y)
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_GA(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
P_IN_AA(x1, x2)  =  P_IN_AA
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)

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

(4) Obligation:

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

MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → P_IN_AA(X, Y)
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_GA(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
P_IN_AA(x1, x2)  =  P_IN_AA
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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

U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))

The TRS R consists of the following rules:

p_in_aa(X, Y) → p_out_aa(X, Y)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

U1_GA(Xs, p_out_aa) → MAP_IN_GA(Xs)
MAP_IN_GA(.(Xs)) → U1_GA(Xs, p_in_aa)

The TRS R consists of the following rules:

p_in_aap_out_aa

The set Q consists of the following terms:

p_in_aa

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

(11) 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:

  • MAP_IN_GA(.(Xs)) → U1_GA(Xs, p_in_aa)
    The graph contains the following edges 1 > 1

  • U1_GA(Xs, p_out_aa) → MAP_IN_GA(Xs)
    The graph contains the following edges 1 >= 1

(12) TRUE

(13) PrologToPiTRSProof (SOUND transformation)

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

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x2, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x1, x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x2, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x1, x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa

(15) 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:

MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → P_IN_AA(X, Y)
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_GA(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x2, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x1, x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
P_IN_AA(x1, x2)  =  P_IN_AA
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x2, x5)

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

(16) Obligation:

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

MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → P_IN_AA(X, Y)
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_GA(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x2, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x1, x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)
P_IN_AA(x1, x2)  =  P_IN_AA
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x2, x5)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(18) Obligation:

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

U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))

The TRS R consists of the following rules:

map_in_ga(.(X, Xs), .(Y, Ys)) → U1_ga(X, Xs, Y, Ys, p_in_aa(X, Y))
p_in_aa(X, Y) → p_out_aa(X, Y)
U1_ga(X, Xs, Y, Ys, p_out_aa(X, Y)) → U2_ga(X, Xs, Y, Ys, map_in_ga(Xs, Ys))
map_in_ga([], []) → map_out_ga([], [])
U2_ga(X, Xs, Y, Ys, map_out_ga(Xs, Ys)) → map_out_ga(.(X, Xs), .(Y, Ys))

The argument filtering Pi contains the following mapping:
map_in_ga(x1, x2)  =  map_in_ga(x1)
.(x1, x2)  =  .(x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x2, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x2, x5)
[]  =  []
map_out_ga(x1, x2)  =  map_out_ga(x1, x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) Obligation:

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

U1_GA(X, Xs, Y, Ys, p_out_aa(X, Y)) → MAP_IN_GA(Xs, Ys)
MAP_IN_GA(.(X, Xs), .(Y, Ys)) → U1_GA(X, Xs, Y, Ys, p_in_aa(X, Y))

The TRS R consists of the following rules:

p_in_aa(X, Y) → p_out_aa(X, Y)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
p_in_aa(x1, x2)  =  p_in_aa
p_out_aa(x1, x2)  =  p_out_aa
MAP_IN_GA(x1, x2)  =  MAP_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x2, x5)

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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) Obligation:

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

U1_GA(Xs, p_out_aa) → MAP_IN_GA(Xs)
MAP_IN_GA(.(Xs)) → U1_GA(Xs, p_in_aa)

The TRS R consists of the following rules:

p_in_aap_out_aa

The set Q consists of the following terms:

p_in_aa

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