/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  a!6220!6220adx : [o] --> o 
  a!6220!6220head : [o] --> o 
  a!6220!6220incr : [o] --> o 
  a!6220!6220nats : [] --> o 
  a!6220!6220tail : [o] --> o 
  a!6220!6220zeros : [] --> o 
  adx : [o] --> o 
  cons : [o * o] --> o 
  head : [o] --> o 
  incr : [o] --> o 
  mark : [o] --> o 
  nats : [] --> o 
  nil : [] --> o 
  s : [o] --> o 
  tail : [o] --> o 
  zeros : [] --> o 

  a!6220!6220incr(nil) => nil 
  a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(nil) => nil 
  a!6220!6220adx(cons(X, Y)) => a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220nats => a!6220!6220adx(a!6220!6220zeros) 
  a!6220!6220zeros => cons(0, zeros) 
  a!6220!6220head(cons(X, Y)) => mark(X) 
  a!6220!6220tail(cons(X, Y)) => mark(Y) 
  mark(incr(X)) => a!6220!6220incr(mark(X)) 
  mark(adx(X)) => a!6220!6220adx(mark(X)) 
  mark(nats) => a!6220!6220nats 
  mark(zeros) => a!6220!6220zeros 
  mark(head(X)) => a!6220!6220head(mark(X)) 
  mark(tail(X)) => a!6220!6220tail(mark(X)) 
  mark(nil) => nil 
  mark(cons(X, Y)) => cons(mark(X), Y) 
  mark(s(X)) => s(mark(X)) 
  mark(0) => 0 
  a!6220!6220incr(X) => incr(X) 
  a!6220!6220adx(X) => adx(X) 
  a!6220!6220nats => nats 
  a!6220!6220zeros => zeros 
  a!6220!6220head(X) => head(X) 
  a!6220!6220tail(X) => tail(X) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  a!6220!6220incr(nil) >? nil 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(nil) >? nil 
  a!6220!6220adx(cons(X, Y)) >? a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220nats >? a!6220!6220adx(a!6220!6220zeros) 
  a!6220!6220zeros >? cons(0, zeros) 
  a!6220!6220head(cons(X, Y)) >? mark(X) 
  a!6220!6220tail(cons(X, Y)) >? mark(Y) 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(adx(X)) >? a!6220!6220adx(mark(X)) 
  mark(nats) >? a!6220!6220nats 
  mark(zeros) >? a!6220!6220zeros 
  mark(head(X)) >? a!6220!6220head(mark(X)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(nil) >? nil 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(s(X)) >? s(mark(X)) 
  mark(0) >? 0 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220adx(X) >? adx(X) 
  a!6220!6220nats >? nats 
  a!6220!6220zeros >? zeros 
  a!6220!6220head(X) >? head(X) 
  a!6220!6220tail(X) >? tail(X) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220adx = \y0.y0 
  a!6220!6220head = \y0.y0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220nats = 0 
  a!6220!6220tail = \y0.1 + y0 
  a!6220!6220zeros = 0 
  adx = \y0.y0 
  cons = \y0y1.2y0 + 2y1 
  head = \y0.y0 
  incr = \y0.y0 
  mark = \y0.y0 
  nats = 0 
  nil = 0 
  s = \y0.y0 
  tail = \y0.1 + y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220incr(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220adx(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220adx(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220incr(cons(mark(_x0), adx(_x1)))]] 
  [[a!6220!6220nats]] = 0 >= 0 = [[a!6220!6220adx(a!6220!6220zeros)]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[cons(0, zeros)]] 
  [[a!6220!6220head(cons(_x0, _x1))]] = 2x0 + 2x1 >= x0 = [[mark(_x0)]] 
  [[a!6220!6220tail(cons(_x0, _x1))]] = 1 + 2x0 + 2x1 > x1 = [[mark(_x1)]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(adx(_x0))]] = x0 >= x0 = [[a!6220!6220adx(mark(_x0))]] 
  [[mark(nats)]] = 0 >= 0 = [[a!6220!6220nats]] 
  [[mark(zeros)]] = 0 >= 0 = [[a!6220!6220zeros]] 
  [[mark(head(_x0))]] = x0 >= x0 = [[a!6220!6220head(mark(_x0))]] 
  [[mark(tail(_x0))]] = 1 + x0 >= 1 + x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220adx(_x0)]] = x0 >= x0 = [[adx(_x0)]] 
  [[a!6220!6220nats]] = 0 >= 0 = [[nats]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[zeros]] 
  [[a!6220!6220head(_x0)]] = x0 >= x0 = [[head(_x0)]] 
  [[a!6220!6220tail(_x0)]] = 1 + x0 >= 1 + x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  a!6220!6220tail(cons(X, Y)) => mark(Y) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  a!6220!6220incr(nil) >? nil 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(nil) >? nil 
  a!6220!6220adx(cons(X, Y)) >? a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220nats >? a!6220!6220adx(a!6220!6220zeros) 
  a!6220!6220zeros >? cons(0, zeros) 
  a!6220!6220head(cons(X, Y)) >? mark(X) 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(adx(X)) >? a!6220!6220adx(mark(X)) 
  mark(nats) >? a!6220!6220nats 
  mark(zeros) >? a!6220!6220zeros 
  mark(head(X)) >? a!6220!6220head(mark(X)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(nil) >? nil 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(s(X)) >? s(mark(X)) 
  mark(0) >? 0 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220adx(X) >? adx(X) 
  a!6220!6220nats >? nats 
  a!6220!6220zeros >? zeros 
  a!6220!6220head(X) >? head(X) 
  a!6220!6220tail(X) >? tail(X) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220adx = \y0.y0 
  a!6220!6220head = \y0.2y0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220nats = 2 
  a!6220!6220tail = \y0.y0 
  a!6220!6220zeros = 0 
  adx = \y0.y0 
  cons = \y0y1.2y0 + 2y1 
  head = \y0.2y0 
  incr = \y0.y0 
  mark = \y0.y0 
  nats = 2 
  nil = 0 
  s = \y0.y0 
  tail = \y0.y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220incr(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220adx(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220adx(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220incr(cons(mark(_x0), adx(_x1)))]] 
  [[a!6220!6220nats]] = 2 > 0 = [[a!6220!6220adx(a!6220!6220zeros)]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[cons(0, zeros)]] 
  [[a!6220!6220head(cons(_x0, _x1))]] = 4x0 + 4x1 >= x0 = [[mark(_x0)]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(adx(_x0))]] = x0 >= x0 = [[a!6220!6220adx(mark(_x0))]] 
  [[mark(nats)]] = 2 >= 2 = [[a!6220!6220nats]] 
  [[mark(zeros)]] = 0 >= 0 = [[a!6220!6220zeros]] 
  [[mark(head(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220head(mark(_x0))]] 
  [[mark(tail(_x0))]] = x0 >= x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220adx(_x0)]] = x0 >= x0 = [[adx(_x0)]] 
  [[a!6220!6220nats]] = 2 >= 2 = [[nats]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[zeros]] 
  [[a!6220!6220head(_x0)]] = 2x0 >= 2x0 = [[head(_x0)]] 
  [[a!6220!6220tail(_x0)]] = x0 >= x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  a!6220!6220nats => a!6220!6220adx(a!6220!6220zeros) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  a!6220!6220incr(nil) >? nil 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(nil) >? nil 
  a!6220!6220adx(cons(X, Y)) >? a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220zeros >? cons(0, zeros) 
  a!6220!6220head(cons(X, Y)) >? mark(X) 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(adx(X)) >? a!6220!6220adx(mark(X)) 
  mark(nats) >? a!6220!6220nats 
  mark(zeros) >? a!6220!6220zeros 
  mark(head(X)) >? a!6220!6220head(mark(X)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(nil) >? nil 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(s(X)) >? s(mark(X)) 
  mark(0) >? 0 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220adx(X) >? adx(X) 
  a!6220!6220nats >? nats 
  a!6220!6220zeros >? zeros 
  a!6220!6220head(X) >? head(X) 
  a!6220!6220tail(X) >? tail(X) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220adx = \y0.y0 
  a!6220!6220head = \y0.2 + y0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220nats = 0 
  a!6220!6220tail = \y0.y0 
  a!6220!6220zeros = 0 
  adx = \y0.y0 
  cons = \y0y1.2y0 + 2y1 
  head = \y0.2 + y0 
  incr = \y0.y0 
  mark = \y0.y0 
  nats = 0 
  nil = 0 
  s = \y0.y0 
  tail = \y0.y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220incr(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220adx(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220adx(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220incr(cons(mark(_x0), adx(_x1)))]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[cons(0, zeros)]] 
  [[a!6220!6220head(cons(_x0, _x1))]] = 2 + 2x0 + 2x1 > x0 = [[mark(_x0)]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(adx(_x0))]] = x0 >= x0 = [[a!6220!6220adx(mark(_x0))]] 
  [[mark(nats)]] = 0 >= 0 = [[a!6220!6220nats]] 
  [[mark(zeros)]] = 0 >= 0 = [[a!6220!6220zeros]] 
  [[mark(head(_x0))]] = 2 + x0 >= 2 + x0 = [[a!6220!6220head(mark(_x0))]] 
  [[mark(tail(_x0))]] = x0 >= x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220adx(_x0)]] = x0 >= x0 = [[adx(_x0)]] 
  [[a!6220!6220nats]] = 0 >= 0 = [[nats]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[zeros]] 
  [[a!6220!6220head(_x0)]] = 2 + x0 >= 2 + x0 = [[head(_x0)]] 
  [[a!6220!6220tail(_x0)]] = x0 >= x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  a!6220!6220head(cons(X, Y)) => mark(X) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  a!6220!6220incr(nil) >? nil 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(nil) >? nil 
  a!6220!6220adx(cons(X, Y)) >? a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220zeros >? cons(0, zeros) 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(adx(X)) >? a!6220!6220adx(mark(X)) 
  mark(nats) >? a!6220!6220nats 
  mark(zeros) >? a!6220!6220zeros 
  mark(head(X)) >? a!6220!6220head(mark(X)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(nil) >? nil 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(s(X)) >? s(mark(X)) 
  mark(0) >? 0 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220adx(X) >? adx(X) 
  a!6220!6220nats >? nats 
  a!6220!6220zeros >? zeros 
  a!6220!6220head(X) >? head(X) 
  a!6220!6220tail(X) >? tail(X) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220adx = \y0.2y0 
  a!6220!6220head = \y0.y0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220nats = 0 
  a!6220!6220tail = \y0.2y0 
  a!6220!6220zeros = 0 
  adx = \y0.2y0 
  cons = \y0y1.y1 + 2y0 
  head = \y0.y0 
  incr = \y0.y0 
  mark = \y0.y0 
  nats = 0 
  nil = 1 
  s = \y0.y0 
  tail = \y0.2y0 
  zeros = 0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220incr(nil)]] = 1 >= 1 = [[nil]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220adx(nil)]] = 2 > 1 = [[nil]] 
  [[a!6220!6220adx(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x0 + 2x1 = [[a!6220!6220incr(cons(mark(_x0), adx(_x1)))]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[cons(0, zeros)]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(adx(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220adx(mark(_x0))]] 
  [[mark(nats)]] = 0 >= 0 = [[a!6220!6220nats]] 
  [[mark(zeros)]] = 0 >= 0 = [[a!6220!6220zeros]] 
  [[mark(head(_x0))]] = x0 >= x0 = [[a!6220!6220head(mark(_x0))]] 
  [[mark(tail(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(nil)]] = 1 >= 1 = [[nil]] 
  [[mark(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220adx(_x0)]] = 2x0 >= 2x0 = [[adx(_x0)]] 
  [[a!6220!6220nats]] = 0 >= 0 = [[nats]] 
  [[a!6220!6220zeros]] = 0 >= 0 = [[zeros]] 
  [[a!6220!6220head(_x0)]] = x0 >= x0 = [[head(_x0)]] 
  [[a!6220!6220tail(_x0)]] = 2x0 >= 2x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  a!6220!6220adx(nil) => nil 

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] a!6220!6220incr#(cons(X, Y)) =#> mark#(X)   
    1] a!6220!6220adx#(cons(X, Y)) =#> a!6220!6220incr#(cons(mark(X), adx(Y)))   
    2] a!6220!6220adx#(cons(X, Y)) =#> mark#(X)   
    3] mark#(incr(X)) =#> a!6220!6220incr#(mark(X))   
    4] mark#(incr(X)) =#> mark#(X)   
    5] mark#(adx(X)) =#> a!6220!6220adx#(mark(X))   
    6] mark#(adx(X)) =#> mark#(X)   
    7] mark#(nats) =#> a!6220!6220nats#   
    8] mark#(zeros) =#> a!6220!6220zeros#   
    9] mark#(head(X)) =#> a!6220!6220head#(mark(X))   
    10] mark#(head(X)) =#> mark#(X)   
    11] mark#(tail(X)) =#> a!6220!6220tail#(mark(X))   
    12] mark#(tail(X)) =#> mark#(X)   
    13] mark#(cons(X, Y)) =#> mark#(X)   
    14] mark#(s(X)) =#> mark#(X)   

  Rules R_0:

    a!6220!6220incr(nil) => nil 
    a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) 
    a!6220!6220adx(cons(X, Y)) => a!6220!6220incr(cons(mark(X), adx(Y))) 
    a!6220!6220zeros => cons(0, zeros) 
    mark(incr(X)) => a!6220!6220incr(mark(X)) 
    mark(adx(X)) => a!6220!6220adx(mark(X)) 
    mark(nats) => a!6220!6220nats 
    mark(zeros) => a!6220!6220zeros 
    mark(head(X)) => a!6220!6220head(mark(X)) 
    mark(tail(X)) => a!6220!6220tail(mark(X)) 
    mark(nil) => nil 
    mark(cons(X, Y)) => cons(mark(X), Y) 
    mark(s(X)) => s(mark(X)) 
    mark(0) => 0 
    a!6220!6220incr(X) => incr(X) 
    a!6220!6220adx(X) => adx(X) 
    a!6220!6220nats => nats 
    a!6220!6220zeros => zeros 
    a!6220!6220head(X) => head(X) 
    a!6220!6220tail(X) => tail(X) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 1 : 0 
    * 2 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 3 : 0 
    * 4 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 5 : 1, 2 
    * 6 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 7 :  
    * 8 :  
    * 9 :  
    * 10 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 11 :  
    * 12 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 13 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 
    * 14 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 

This graph has the following strongly connected components:

  P_1:

    a!6220!6220incr#(cons(X, Y)) =#> mark#(X)   
    a!6220!6220adx#(cons(X, Y)) =#> a!6220!6220incr#(cons(mark(X), adx(Y)))   
    a!6220!6220adx#(cons(X, Y)) =#> mark#(X)   
    mark#(incr(X)) =#> a!6220!6220incr#(mark(X))   
    mark#(incr(X)) =#> mark#(X)   
    mark#(adx(X)) =#> a!6220!6220adx#(mark(X))   
    mark#(adx(X)) =#> mark#(X)   
    mark#(head(X)) =#> mark#(X)   
    mark#(tail(X)) =#> mark#(X)   
    mark#(cons(X, Y)) =#> mark#(X)   
    mark#(s(X)) =#> mark#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

The formative rules of (P_1, R_0) are R_1 ::=

  a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(cons(X, Y)) => a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220zeros => cons(0, zeros) 
  mark(incr(X)) => a!6220!6220incr(mark(X)) 
  mark(adx(X)) => a!6220!6220adx(mark(X)) 
  mark(zeros) => a!6220!6220zeros 
  mark(head(X)) => a!6220!6220head(mark(X)) 
  mark(tail(X)) => a!6220!6220tail(mark(X)) 
  mark(cons(X, Y)) => cons(mark(X), Y) 
  mark(s(X)) => s(mark(X)) 
  a!6220!6220incr(X) => incr(X) 
  a!6220!6220adx(X) => adx(X) 
  a!6220!6220zeros => zeros 
  a!6220!6220head(X) => head(X) 
  a!6220!6220tail(X) => tail(X) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  a!6220!6220incr#(cons(X, Y)) >? mark#(X) 
  a!6220!6220adx#(cons(X, Y)) >? a!6220!6220incr#(cons(mark(X), adx(Y))) 
  a!6220!6220adx#(cons(X, Y)) >? mark#(X) 
  mark#(incr(X)) >? a!6220!6220incr#(mark(X)) 
  mark#(incr(X)) >? mark#(X) 
  mark#(adx(X)) >? a!6220!6220adx#(mark(X)) 
  mark#(adx(X)) >? mark#(X) 
  mark#(head(X)) >? mark#(X) 
  mark#(tail(X)) >? mark#(X) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  a!6220!6220incr(cons(X, Y)) >= cons(s(mark(X)), incr(Y)) 
  a!6220!6220adx(cons(X, Y)) >= a!6220!6220incr(cons(mark(X), adx(Y))) 
  a!6220!6220zeros >= cons(0, zeros) 
  mark(incr(X)) >= a!6220!6220incr(mark(X)) 
  mark(adx(X)) >= a!6220!6220adx(mark(X)) 
  mark(zeros) >= a!6220!6220zeros 
  mark(head(X)) >= a!6220!6220head(mark(X)) 
  mark(tail(X)) >= a!6220!6220tail(mark(X)) 
  mark(cons(X, Y)) >= cons(mark(X), Y) 
  mark(s(X)) >= s(mark(X)) 
  a!6220!6220incr(X) >= incr(X) 
  a!6220!6220adx(X) >= adx(X) 
  a!6220!6220zeros >= zeros 
  a!6220!6220head(X) >= head(X) 
  a!6220!6220tail(X) >= tail(X) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220adx = \y0.y0 
  a!6220!6220adx# = \y0.y0 
  a!6220!6220head = \y0.y0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220incr# = \y0.y0 
  a!6220!6220tail = \y0.2y0 
  a!6220!6220zeros = 2 
  adx = \y0.y0 
  cons = \y0y1.2 + 2y0 
  head = \y0.y0 
  incr = \y0.y0 
  mark = \y0.y0 
  mark# = \y0.1 + 2y0 
  s = \y0.y0 
  tail = \y0.2y0 
  zeros = 2 

Using this interpretation, the requirements translate to:

  [[a!6220!6220incr#(cons(_x0, _x1))]] = 2 + 2x0 > 1 + 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220adx#(cons(_x0, _x1))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220incr#(cons(mark(_x0), adx(_x1)))]] 
  [[a!6220!6220adx#(cons(_x0, _x1))]] = 2 + 2x0 > 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 1 + 2x0 > x0 = [[a!6220!6220incr#(mark(_x0))]] 
  [[mark#(incr(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(adx(_x0))]] = 1 + 2x0 > x0 = [[a!6220!6220adx#(mark(_x0))]] 
  [[mark#(adx(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(head(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(tail(_x0))]] = 1 + 4x0 >= 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(cons(_x0, _x1))]] = 5 + 4x0 > 1 + 2x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2 + 2x0 >= 2 + 2x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220adx(cons(_x0, _x1))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220incr(cons(mark(_x0), adx(_x1)))]] 
  [[a!6220!6220zeros]] = 2 >= 2 = [[cons(0, zeros)]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(adx(_x0))]] = x0 >= x0 = [[a!6220!6220adx(mark(_x0))]] 
  [[mark(zeros)]] = 2 >= 2 = [[a!6220!6220zeros]] 
  [[mark(head(_x0))]] = x0 >= x0 = [[a!6220!6220head(mark(_x0))]] 
  [[mark(tail(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = 2 + 2x0 >= 2 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220adx(_x0)]] = x0 >= x0 = [[adx(_x0)]] 
  [[a!6220!6220zeros]] = 2 >= 2 = [[zeros]] 
  [[a!6220!6220head(_x0)]] = x0 >= x0 = [[head(_x0)]] 
  [[a!6220!6220tail(_x0)]] = 2x0 >= 2x0 = [[tail(_x0)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of:

  a!6220!6220adx#(cons(X, Y)) =#> a!6220!6220incr#(cons(mark(X), adx(Y)))   
  mark#(incr(X)) =#> mark#(X)   
  mark#(adx(X)) =#> mark#(X)   
  mark#(head(X)) =#> mark#(X)   
  mark#(tail(X)) =#> mark#(X)   
  mark#(s(X)) =#> mark#(X)   

Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_1, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 : 1, 2, 3, 4, 5 
    * 2 : 1, 2, 3, 4, 5 
    * 3 : 1, 2, 3, 4, 5 
    * 4 : 1, 2, 3, 4, 5 
    * 5 : 1, 2, 3, 4, 5 

This graph has the following strongly connected components:

  P_3:

    mark#(incr(X)) =#> mark#(X)   
    mark#(adx(X)) =#> mark#(X)   
    mark#(head(X)) =#> mark#(X)   
    mark#(tail(X)) =#> mark#(X)   
    mark#(s(X)) =#> mark#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_2, R_1, m, f) by (P_3, R_1, m, f).

Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(mark#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(mark#(incr(X))) = incr(X) |> X = nu(mark#(X)) 
  nu(mark#(adx(X))) = adx(X) |> X = nu(mark#(X)) 
  nu(mark#(head(X))) = head(X) |> X = nu(mark#(X)) 
  nu(mark#(tail(X))) = tail(X) |> X = nu(mark#(X)) 
  nu(mark#(s(X))) = s(X) |> X = nu(mark#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_1, minimal, f) by ({}, R_1, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.