(0) Obligation:

Clauses:

countstack(empty, 0).
countstack(push(nil, T), X) :- countstack(T, X).
countstack(push(cons(U, V), T), s(X)) :- countstack(push(U, push(V, T)), X).

Query: countstack(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

(2) Obligation:

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

countstackA_in_ga(empty, 0) → countstackA_out_ga(empty, 0)
countstackA_in_ga(push(nil, empty), 0) → countstackA_out_ga(push(nil, empty), 0)
countstackA_in_ga(push(nil, push(nil, T16)), T18) → U1_ga(T16, T18, countstackA_in_ga(T16, T18))
countstackA_in_ga(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_ga(T35, T36, T37, T39, countstackA_in_ga(push(T35, push(T36, T37)), T39))
countstackA_in_ga(push(cons(nil, T61), T62), s(T64)) → U3_ga(T61, T62, T64, countstackA_in_ga(push(T61, T62), T64))
countstackA_in_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → U4_ga(T75, T76, T77, T78, T80, countstackA_in_ga(push(T75, push(T76, push(T77, T78))), T80))
U4_ga(T75, T76, T77, T78, T80, countstackA_out_ga(push(T75, push(T76, push(T77, T78))), T80)) → countstackA_out_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80)))
U3_ga(T61, T62, T64, countstackA_out_ga(push(T61, T62), T64)) → countstackA_out_ga(push(cons(nil, T61), T62), s(T64))
U2_ga(T35, T36, T37, T39, countstackA_out_ga(push(T35, push(T36, T37)), T39)) → countstackA_out_ga(push(nil, push(cons(T35, T36), T37)), s(T39))
U1_ga(T16, T18, countstackA_out_ga(T16, T18)) → countstackA_out_ga(push(nil, push(nil, T16)), T18)

The argument filtering Pi contains the following mapping:
countstackA_in_ga(x1, x2)  =  countstackA_in_ga(x1)
empty  =  empty
countstackA_out_ga(x1, x2)  =  countstackA_out_ga(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x1, x2, x4)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x1, x2, x3, x4, x6)
s(x1)  =  s(x1)

(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:

COUNTSTACKA_IN_GA(push(nil, push(nil, T16)), T18) → U1_GA(T16, T18, countstackA_in_ga(T16, T18))
COUNTSTACKA_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACKA_IN_GA(T16, T18)
COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_GA(T35, T36, T37, T39, countstackA_in_ga(push(T35, push(T36, T37)), T39))
COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACKA_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62), s(T64)) → U3_GA(T61, T62, T64, countstackA_in_ga(push(T61, T62), T64))
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62), s(T64)) → COUNTSTACKA_IN_GA(push(T61, T62), T64)
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → U4_GA(T75, T76, T77, T78, T80, countstackA_in_ga(push(T75, push(T76, push(T77, T78))), T80))
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → COUNTSTACKA_IN_GA(push(T75, push(T76, push(T77, T78))), T80)

The TRS R consists of the following rules:

countstackA_in_ga(empty, 0) → countstackA_out_ga(empty, 0)
countstackA_in_ga(push(nil, empty), 0) → countstackA_out_ga(push(nil, empty), 0)
countstackA_in_ga(push(nil, push(nil, T16)), T18) → U1_ga(T16, T18, countstackA_in_ga(T16, T18))
countstackA_in_ga(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_ga(T35, T36, T37, T39, countstackA_in_ga(push(T35, push(T36, T37)), T39))
countstackA_in_ga(push(cons(nil, T61), T62), s(T64)) → U3_ga(T61, T62, T64, countstackA_in_ga(push(T61, T62), T64))
countstackA_in_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → U4_ga(T75, T76, T77, T78, T80, countstackA_in_ga(push(T75, push(T76, push(T77, T78))), T80))
U4_ga(T75, T76, T77, T78, T80, countstackA_out_ga(push(T75, push(T76, push(T77, T78))), T80)) → countstackA_out_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80)))
U3_ga(T61, T62, T64, countstackA_out_ga(push(T61, T62), T64)) → countstackA_out_ga(push(cons(nil, T61), T62), s(T64))
U2_ga(T35, T36, T37, T39, countstackA_out_ga(push(T35, push(T36, T37)), T39)) → countstackA_out_ga(push(nil, push(cons(T35, T36), T37)), s(T39))
U1_ga(T16, T18, countstackA_out_ga(T16, T18)) → countstackA_out_ga(push(nil, push(nil, T16)), T18)

The argument filtering Pi contains the following mapping:
countstackA_in_ga(x1, x2)  =  countstackA_in_ga(x1)
empty  =  empty
countstackA_out_ga(x1, x2)  =  countstackA_out_ga(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x1, x2, x4)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x1, x2, x3, x4, x6)
s(x1)  =  s(x1)
COUNTSTACKA_IN_GA(x1, x2)  =  COUNTSTACKA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x1, x2, x4)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x1, x2, x3, x4, x6)

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

(4) Obligation:

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

COUNTSTACKA_IN_GA(push(nil, push(nil, T16)), T18) → U1_GA(T16, T18, countstackA_in_ga(T16, T18))
COUNTSTACKA_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACKA_IN_GA(T16, T18)
COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_GA(T35, T36, T37, T39, countstackA_in_ga(push(T35, push(T36, T37)), T39))
COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACKA_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62), s(T64)) → U3_GA(T61, T62, T64, countstackA_in_ga(push(T61, T62), T64))
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62), s(T64)) → COUNTSTACKA_IN_GA(push(T61, T62), T64)
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → U4_GA(T75, T76, T77, T78, T80, countstackA_in_ga(push(T75, push(T76, push(T77, T78))), T80))
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → COUNTSTACKA_IN_GA(push(T75, push(T76, push(T77, T78))), T80)

The TRS R consists of the following rules:

countstackA_in_ga(empty, 0) → countstackA_out_ga(empty, 0)
countstackA_in_ga(push(nil, empty), 0) → countstackA_out_ga(push(nil, empty), 0)
countstackA_in_ga(push(nil, push(nil, T16)), T18) → U1_ga(T16, T18, countstackA_in_ga(T16, T18))
countstackA_in_ga(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_ga(T35, T36, T37, T39, countstackA_in_ga(push(T35, push(T36, T37)), T39))
countstackA_in_ga(push(cons(nil, T61), T62), s(T64)) → U3_ga(T61, T62, T64, countstackA_in_ga(push(T61, T62), T64))
countstackA_in_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → U4_ga(T75, T76, T77, T78, T80, countstackA_in_ga(push(T75, push(T76, push(T77, T78))), T80))
U4_ga(T75, T76, T77, T78, T80, countstackA_out_ga(push(T75, push(T76, push(T77, T78))), T80)) → countstackA_out_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80)))
U3_ga(T61, T62, T64, countstackA_out_ga(push(T61, T62), T64)) → countstackA_out_ga(push(cons(nil, T61), T62), s(T64))
U2_ga(T35, T36, T37, T39, countstackA_out_ga(push(T35, push(T36, T37)), T39)) → countstackA_out_ga(push(nil, push(cons(T35, T36), T37)), s(T39))
U1_ga(T16, T18, countstackA_out_ga(T16, T18)) → countstackA_out_ga(push(nil, push(nil, T16)), T18)

The argument filtering Pi contains the following mapping:
countstackA_in_ga(x1, x2)  =  countstackA_in_ga(x1)
empty  =  empty
countstackA_out_ga(x1, x2)  =  countstackA_out_ga(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x1, x2, x4)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x1, x2, x3, x4, x6)
s(x1)  =  s(x1)
COUNTSTACKA_IN_GA(x1, x2)  =  COUNTSTACKA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x1, x2, x4)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x1, x2, x3, x4, x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACKA_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACKA_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACKA_IN_GA(T16, T18)
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62), s(T64)) → COUNTSTACKA_IN_GA(push(T61, T62), T64)
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → COUNTSTACKA_IN_GA(push(T75, push(T76, push(T77, T78))), T80)

The TRS R consists of the following rules:

countstackA_in_ga(empty, 0) → countstackA_out_ga(empty, 0)
countstackA_in_ga(push(nil, empty), 0) → countstackA_out_ga(push(nil, empty), 0)
countstackA_in_ga(push(nil, push(nil, T16)), T18) → U1_ga(T16, T18, countstackA_in_ga(T16, T18))
countstackA_in_ga(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_ga(T35, T36, T37, T39, countstackA_in_ga(push(T35, push(T36, T37)), T39))
countstackA_in_ga(push(cons(nil, T61), T62), s(T64)) → U3_ga(T61, T62, T64, countstackA_in_ga(push(T61, T62), T64))
countstackA_in_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → U4_ga(T75, T76, T77, T78, T80, countstackA_in_ga(push(T75, push(T76, push(T77, T78))), T80))
U4_ga(T75, T76, T77, T78, T80, countstackA_out_ga(push(T75, push(T76, push(T77, T78))), T80)) → countstackA_out_ga(push(cons(cons(T75, T76), T77), T78), s(s(T80)))
U3_ga(T61, T62, T64, countstackA_out_ga(push(T61, T62), T64)) → countstackA_out_ga(push(cons(nil, T61), T62), s(T64))
U2_ga(T35, T36, T37, T39, countstackA_out_ga(push(T35, push(T36, T37)), T39)) → countstackA_out_ga(push(nil, push(cons(T35, T36), T37)), s(T39))
U1_ga(T16, T18, countstackA_out_ga(T16, T18)) → countstackA_out_ga(push(nil, push(nil, T16)), T18)

The argument filtering Pi contains the following mapping:
countstackA_in_ga(x1, x2)  =  countstackA_in_ga(x1)
empty  =  empty
countstackA_out_ga(x1, x2)  =  countstackA_out_ga(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
U3_ga(x1, x2, x3, x4)  =  U3_ga(x1, x2, x4)
U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x1, x2, x3, x4, x6)
s(x1)  =  s(x1)
COUNTSTACKA_IN_GA(x1, x2)  =  COUNTSTACKA_IN_GA(x1)

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:

COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACKA_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACKA_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACKA_IN_GA(T16, T18)
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62), s(T64)) → COUNTSTACKA_IN_GA(push(T61, T62), T64)
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78), s(s(T80))) → COUNTSTACKA_IN_GA(push(T75, push(T76, push(T77, T78))), T80)

R is empty.
The argument filtering Pi contains the following mapping:
push(x1, x2)  =  push(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
COUNTSTACKA_IN_GA(x1, x2)  =  COUNTSTACKA_IN_GA(x1)

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:

COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37))) → COUNTSTACKA_IN_GA(push(T35, push(T36, T37)))
COUNTSTACKA_IN_GA(push(nil, push(nil, T16))) → COUNTSTACKA_IN_GA(T16)
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62)) → COUNTSTACKA_IN_GA(push(T61, T62))
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78)) → COUNTSTACKA_IN_GA(push(T75, push(T76, push(T77, T78))))

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

(11) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

COUNTSTACKA_IN_GA(push(nil, push(cons(T35, T36), T37))) → COUNTSTACKA_IN_GA(push(T35, push(T36, T37)))
COUNTSTACKA_IN_GA(push(nil, push(nil, T16))) → COUNTSTACKA_IN_GA(T16)
COUNTSTACKA_IN_GA(push(cons(nil, T61), T62)) → COUNTSTACKA_IN_GA(push(T61, T62))
COUNTSTACKA_IN_GA(push(cons(cons(T75, T76), T77), T78)) → COUNTSTACKA_IN_GA(push(T75, push(T76, push(T77, T78))))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(COUNTSTACKA_IN_GA(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(push(x1, x2)) = 2·x1 + x2   

(12) Obligation:

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) YES