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


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YES
We consider the system theBenchmark.

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

  0 : [] --> o 
  a!6220!6220incr : [o] --> o 
  a!6220!6220oddNs : [] --> o 
  a!6220!6220pairNs : [] --> o 
  a!6220!6220repItems : [o] --> o 
  a!6220!6220tail : [o] --> o 
  a!6220!6220take : [o * o] --> o 
  a!6220!6220zip : [o * o] --> o 
  cons : [o * o] --> o 
  incr : [o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  oddNs : [] --> o 
  pair : [o * o] --> o 
  pairNs : [] --> o 
  repItems : [o] --> o 
  s : [o] --> o 
  tail : [o] --> o 
  take : [o * o] --> o 
  zip : [o * o] --> o 

  a!6220!6220pairNs => cons(0, incr(oddNs)) 
  a!6220!6220oddNs => a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(0, X) => nil 
  a!6220!6220take(s(X), cons(Y, Z)) => cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(nil, X) => nil 
  a!6220!6220zip(X, nil) => nil 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) => cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220tail(cons(X, Y)) => mark(Y) 
  a!6220!6220repItems(nil) => nil 
  a!6220!6220repItems(cons(X, Y)) => cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) => a!6220!6220pairNs 
  mark(incr(X)) => a!6220!6220incr(mark(X)) 
  mark(oddNs) => a!6220!6220oddNs 
  mark(take(X, Y)) => a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) => a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) => a!6220!6220tail(mark(X)) 
  mark(repItems(X)) => a!6220!6220repItems(mark(X)) 
  mark(cons(X, Y)) => cons(mark(X), Y) 
  mark(0) => 0 
  mark(s(X)) => s(mark(X)) 
  mark(nil) => nil 
  mark(pair(X, Y)) => pair(mark(X), mark(Y)) 
  a!6220!6220pairNs => pairNs 
  a!6220!6220incr(X) => incr(X) 
  a!6220!6220oddNs => oddNs 
  a!6220!6220take(X, Y) => take(X, Y) 
  a!6220!6220zip(X, Y) => zip(X, Y) 
  a!6220!6220tail(X) => tail(X) 
  a!6220!6220repItems(X) => repItems(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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(0, X) >? nil 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(nil, X) >? nil 
  a!6220!6220zip(X, nil) >? nil 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220tail(cons(X, Y)) >? mark(Y) 
  a!6220!6220repItems(nil) >? nil 
  a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(repItems(X)) >? a!6220!6220repItems(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(nil) >? nil 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  a!6220!6220tail(X) >? tail(X) 
  a!6220!6220repItems(X) >? repItems(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220repItems = \y0.2y0 
  a!6220!6220tail = \y0.y0 
  a!6220!6220take = \y0y1.y0 + 2y1 
  a!6220!6220zip = \y0y1.1 + y0 + y1 
  cons = \y0y1.y1 + 2y0 
  incr = \y0.2y0 
  mark = \y0.y0 
  nil = 0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  repItems = \y0.2y0 
  s = \y0.y0 
  tail = \y0.y0 
  take = \y0y1.y0 + 2y1 
  zip = \y0y1.1 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(0, _x0)]] = 2x0 >= 0 = [[nil]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = x0 + 2x2 + 4x1 >= x0 + 2x1 + 2x2 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(nil, _x0)]] = 1 + x0 > 0 = [[nil]] 
  [[a!6220!6220zip(_x0, nil)]] = 1 + x0 > 0 = [[nil]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 1 + x1 + x3 + 2x0 + 2x2 >= 1 + x1 + x3 + 2x0 + 2x2 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[a!6220!6220tail(cons(_x0, _x1))]] = x1 + 2x0 >= x1 = [[mark(_x1)]] 
  [[a!6220!6220repItems(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220repItems(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[cons(mark(_x0), cons(_x0, repItems(_x1)))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = x0 >= x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(repItems(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220repItems(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(pair(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = x0 >= x0 = [[tail(_x0)]] 
  [[a!6220!6220repItems(_x0)]] = 2x0 >= 2x0 = [[repItems(_x0)]] 

We can thus remove the following rules:

  a!6220!6220zip(nil, X) => nil 
  a!6220!6220zip(X, nil) => nil 

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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(0, X) >? nil 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220tail(cons(X, Y)) >? mark(Y) 
  a!6220!6220repItems(nil) >? nil 
  a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(repItems(X)) >? a!6220!6220repItems(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(nil) >? nil 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  a!6220!6220tail(X) >? tail(X) 
  a!6220!6220repItems(X) >? repItems(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220repItems = \y0.2y0 
  a!6220!6220tail = \y0.2 + y0 
  a!6220!6220take = \y0y1.y0 + y1 
  a!6220!6220zip = \y0y1.y0 + y1 
  cons = \y0y1.y0 + y1 
  incr = \y0.y0 
  mark = \y0.y0 
  nil = 0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  repItems = \y0.2y0 
  s = \y0.y0 
  tail = \y0.2 + y0 
  take = \y0y1.y0 + y1 
  zip = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(0, _x0)]] = x0 >= 0 = [[nil]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = x0 + x1 + x2 + x3 >= x0 + x1 + x2 + x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[a!6220!6220tail(cons(_x0, _x1))]] = 2 + x0 + x1 > x1 = [[mark(_x1)]] 
  [[a!6220!6220repItems(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220repItems(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(mark(_x0), cons(_x0, repItems(_x1)))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(repItems(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220repItems(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(pair(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = 2 + x0 >= 2 + x0 = [[tail(_x0)]] 
  [[a!6220!6220repItems(_x0)]] = 2x0 >= 2x0 = [[repItems(_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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(0, X) >? nil 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220repItems(nil) >? nil 
  a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(repItems(X)) >? a!6220!6220repItems(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(nil) >? nil 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  a!6220!6220tail(X) >? tail(X) 
  a!6220!6220repItems(X) >? repItems(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220repItems = \y0.2y0 
  a!6220!6220tail = \y0.y0 
  a!6220!6220take = \y0y1.1 + y1 + 2y0 
  a!6220!6220zip = \y0y1.y0 + 2y1 
  cons = \y0y1.y1 + 2y0 
  incr = \y0.2y0 
  mark = \y0.y0 
  nil = 0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  repItems = \y0.2y0 
  s = \y0.2y0 
  tail = \y0.y0 
  take = \y0y1.1 + y1 + 2y0 
  zip = \y0y1.y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(0, _x0)]] = 1 + x0 > 0 = [[nil]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = 1 + x2 + 2x1 + 4x0 >= 1 + x2 + 2x0 + 2x1 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = x1 + 2x0 + 2x3 + 4x2 >= x1 + 2x0 + 2x2 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[a!6220!6220repItems(nil)]] = 0 >= 0 = [[nil]] 
  [[a!6220!6220repItems(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[cons(mark(_x0), cons(_x0, repItems(_x1)))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = x0 >= x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(repItems(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220repItems(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(nil)]] = 0 >= 0 = [[nil]] 
  [[mark(pair(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = x0 >= x0 = [[tail(_x0)]] 
  [[a!6220!6220repItems(_x0)]] = 2x0 >= 2x0 = [[repItems(_x0)]] 

We can thus remove the following rules:

  a!6220!6220take(0, X) => nil 

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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220repItems(nil) >? nil 
  a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(repItems(X)) >? a!6220!6220repItems(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(nil) >? nil 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  a!6220!6220tail(X) >? tail(X) 
  a!6220!6220repItems(X) >? repItems(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220repItems = \y0.2y0 
  a!6220!6220tail = \y0.y0 
  a!6220!6220take = \y0y1.y0 + y1 
  a!6220!6220zip = \y0y1.y0 + 2y1 
  cons = \y0y1.y0 + y1 
  incr = \y0.y0 
  mark = \y0.y0 
  nil = 2 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  repItems = \y0.2y0 
  s = \y0.y0 
  tail = \y0.y0 
  take = \y0y1.y0 + y1 
  zip = \y0y1.y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = x0 + x1 + 2x2 + 2x3 >= x0 + x1 + x2 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[a!6220!6220repItems(nil)]] = 4 > 2 = [[nil]] 
  [[a!6220!6220repItems(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(mark(_x0), cons(_x0, repItems(_x1)))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = x0 >= x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = x0 >= x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(repItems(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220repItems(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(nil)]] = 2 >= 2 = [[nil]] 
  [[mark(pair(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = x0 >= x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = x0 >= x0 = [[tail(_x0)]] 
  [[a!6220!6220repItems(_x0)]] = 2x0 >= 2x0 = [[repItems(_x0)]] 

We can thus remove the following rules:

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

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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(repItems(X)) >? a!6220!6220repItems(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(nil) >? nil 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  a!6220!6220tail(X) >? tail(X) 
  a!6220!6220repItems(X) >? repItems(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220repItems = \y0.2 + 3y0 
  a!6220!6220tail = \y0.y0 
  a!6220!6220take = \y0y1.y0 + 2y1 
  a!6220!6220zip = \y0y1.2y0 + 2y1 
  cons = \y0y1.y1 + 2y0 
  incr = \y0.2y0 
  mark = \y0.2y0 
  nil = 2 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  repItems = \y0.2 + 3y0 
  s = \y0.y0 
  tail = \y0.y0 
  take = \y0y1.y0 + 2y1 
  zip = \y0y1.2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = x0 + 2x2 + 4x1 >= x0 + 2x2 + 4x1 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2x1 + 2x3 + 4x0 + 4x2 >= 2x1 + 2x3 + 4x0 + 4x2 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[a!6220!6220repItems(cons(_x0, _x1))]] = 2 + 3x1 + 6x0 >= 2 + 3x1 + 6x0 = [[cons(mark(_x0), cons(_x0, repItems(_x1)))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(repItems(_x0))]] = 4 + 6x0 > 2 + 6x0 = [[a!6220!6220repItems(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = 2x1 + 4x0 >= x1 + 4x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(nil)]] = 4 > 2 = [[nil]] 
  [[mark(pair(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = x0 >= x0 = [[tail(_x0)]] 
  [[a!6220!6220repItems(_x0)]] = 2 + 3x0 >= 2 + 3x0 = [[repItems(_x0)]] 

We can thus remove the following rules:

  mark(repItems(X)) => a!6220!6220repItems(mark(X)) 
  mark(nil) => nil 

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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  a!6220!6220tail(X) >? tail(X) 
  a!6220!6220repItems(X) >? repItems(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220repItems = \y0.3 + 3y0 
  a!6220!6220tail = \y0.2y0 
  a!6220!6220take = \y0y1.y0 + 2y1 
  a!6220!6220zip = \y0y1.2y0 + 2y1 
  cons = \y0y1.y1 + 2y0 
  incr = \y0.2y0 
  mark = \y0.2y0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  repItems = \y0.y0 
  s = \y0.y0 
  tail = \y0.2y0 
  take = \y0y1.y0 + 2y1 
  zip = \y0y1.2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = x0 + 2x2 + 4x1 >= x0 + 2x2 + 4x1 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2x1 + 2x3 + 4x0 + 4x2 >= 2x1 + 2x3 + 4x0 + 4x2 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[a!6220!6220repItems(cons(_x0, _x1))]] = 3 + 3x1 + 6x0 > x1 + 6x0 = [[cons(mark(_x0), cons(_x0, repItems(_x1)))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = 2x1 + 4x0 >= x1 + 4x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(pair(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = 2x0 >= 2x0 = [[tail(_x0)]] 
  [[a!6220!6220repItems(_x0)]] = 3 + 3x0 > x0 = [[repItems(_x0)]] 

We can thus remove the following rules:

  a!6220!6220repItems(cons(X, Y)) => cons(mark(X), cons(X, repItems(Y))) 
  a!6220!6220repItems(X) => repItems(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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(tail(X)) >? a!6220!6220tail(mark(X)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  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!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220tail = \y0.1 + y0 
  a!6220!6220take = \y0y1.2y0 + 2y1 
  a!6220!6220zip = \y0y1.2y0 + 2y1 
  cons = \y0y1.y0 + y1 
  incr = \y0.2y0 
  mark = \y0.2y0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  s = \y0.y0 
  tail = \y0.1 + y0 
  take = \y0y1.2y0 + 2y1 
  zip = \y0y1.2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2x0 + 2x1 + 2x2 + 2x3 >= 2x0 + 2x1 + 2x2 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(tail(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[a!6220!6220tail(mark(_x0))]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(pair(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = 1 + x0 >= 1 + x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  mark(tail(X)) => a!6220!6220tail(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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 
  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!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220tail = \y0.3 + y0 
  a!6220!6220take = \y0y1.y0 + y1 
  a!6220!6220zip = \y0y1.y0 + y1 
  cons = \y0y1.y0 + 2y1 
  incr = \y0.2y0 
  mark = \y0.y0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  s = \y0.2y0 
  tail = \y0.y0 
  take = \y0y1.y0 + y1 
  zip = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = x1 + 2x0 + 2x2 >= x1 + 2x0 + 2x2 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = x0 + x2 + 2x1 + 2x3 >= x0 + x2 + 2x1 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(cons(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(pair(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[zip(_x0, _x1)]] 
  [[a!6220!6220tail(_x0)]] = 3 + x0 > x0 = [[tail(_x0)]] 

We can thus remove the following rules:

  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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220take(X, Y) >? take(X, Y) 
  a!6220!6220zip(X, Y) >? zip(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220take = \y0y1.2 + y0 + 2y1 
  a!6220!6220zip = \y0y1.2y0 + 2y1 
  cons = \y0y1.y0 + y1 
  incr = \y0.2y0 
  mark = \y0.2y0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  s = \y0.y0 
  take = \y0y1.1 + y0 + 2y1 
  zip = \y0y1.2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220take(s(_x0), cons(_x1, _x2))]] = 2 + x0 + 2x1 + 2x2 > 1 + x0 + 2x1 + 2x2 = [[cons(mark(_x1), take(_x0, _x2))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2x0 + 2x1 + 2x2 + 2x3 >= 2x0 + 2x1 + 2x2 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = 2 + 2x0 + 4x1 >= 2 + 2x0 + 4x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(pair(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220take(_x0, _x1)]] = 2 + x0 + 2x1 > 1 + x0 + 2x1 = [[take(_x0, _x1)]] 
  [[a!6220!6220zip(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[zip(_x0, _x1)]] 

We can thus remove the following rules:

  a!6220!6220take(s(X), cons(Y, Z)) => cons(mark(Y), take(X, Z)) 
  a!6220!6220take(X, Y) => take(X, 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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220zip(X, Y) >? zip(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220take = \y0y1.y0 + y1 
  a!6220!6220zip = \y0y1.y0 + y1 
  cons = \y0y1.y0 + 2y1 
  incr = \y0.2y0 
  mark = \y0.y0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  s = \y0.y0 
  take = \y0y1.3 + 3y0 + 3y1 
  zip = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 4x1 >= x0 + 4x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = x0 + x2 + 2x1 + 2x3 >= x0 + x2 + 2x1 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(take(_x0, _x1))]] = 3 + 3x0 + 3x1 > x0 + x1 = [[a!6220!6220take(mark(_x0), mark(_x1))]] 
  [[mark(zip(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(cons(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] 
  [[mark(pair(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220zip(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[zip(_x0, _x1)]] 

We can thus remove the following rules:

  mark(take(X, Y)) => a!6220!6220take(mark(X), 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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  mark(pair(X, Y)) >? pair(mark(X), mark(Y)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 
  a!6220!6220zip(X, Y) >? zip(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220zip = \y0y1.2 + 2y0 + 2y1 
  cons = \y0y1.y0 + y1 
  incr = \y0.2y0 
  mark = \y0.2y0 
  oddNs = 0 
  pair = \y0y1.1 + y0 + y1 
  pairNs = 0 
  s = \y0.y0 
  zip = \y0y1.1 + 2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2 + 2x0 + 2x1 + 2x2 + 2x3 >= 2 + 2x0 + 2x1 + 2x2 + 2x3 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(zip(_x0, _x1))]] = 2 + 4x0 + 4x1 >= 2 + 4x0 + 4x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[mark(pair(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[pair(mark(_x0), mark(_x1))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 
  [[a!6220!6220zip(_x0, _x1)]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[zip(_x0, _x1)]] 

We can thus remove the following rules:

  mark(pair(X, Y)) => pair(mark(X), mark(Y)) 
  a!6220!6220zip(X, Y) => zip(X, 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!6220pairNs >? cons(0, incr(oddNs)) 
  a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) 
  a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(pairNs) >? a!6220!6220pairNs 
  mark(incr(X)) >? a!6220!6220incr(mark(X)) 
  mark(oddNs) >? a!6220!6220oddNs 
  mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) 
  mark(cons(X, Y)) >? cons(mark(X), Y) 
  mark(0) >? 0 
  mark(s(X)) >? s(mark(X)) 
  a!6220!6220pairNs >? pairNs 
  a!6220!6220incr(X) >? incr(X) 
  a!6220!6220oddNs >? oddNs 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220oddNs = 0 
  a!6220!6220pairNs = 0 
  a!6220!6220zip = \y0y1.3 + 2y0 + 2y1 
  cons = \y0y1.y1 + 2y0 
  incr = \y0.2y0 
  mark = \y0.2y0 
  oddNs = 0 
  pair = \y0y1.y0 + y1 
  pairNs = 0 
  s = \y0.y0 
  zip = \y0y1.2 + 2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[a!6220!6220zip(cons(_x0, _x1), cons(_x2, _x3))]] = 3 + 2x1 + 2x3 + 4x0 + 4x2 > 2 + 2x1 + 2x3 + 4x0 + 4x2 = [[cons(pair(mark(_x0), mark(_x2)), zip(_x1, _x3))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 0 >= 0 = [[a!6220!6220oddNs]] 
  [[mark(zip(_x0, _x1))]] = 4 + 4x0 + 4x1 > 3 + 4x0 + 4x1 = [[a!6220!6220zip(mark(_x0), mark(_x1))]] 
  [[mark(cons(_x0, _x1))]] = 2x1 + 4x0 >= x1 + 4x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(0)]] = 0 >= 0 = [[0]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 0 >= 0 = [[oddNs]] 

We can thus remove the following rules:

  a!6220!6220zip(cons(X, Y), cons(Z, U)) => cons(pair(mark(X), mark(Z)), zip(Y, U)) 
  mark(zip(X, Y)) => a!6220!6220zip(mark(X), mark(Y)) 

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!6220oddNs# =#> a!6220!6220incr#(a!6220!6220pairNs)   
    1] a!6220!6220oddNs# =#> a!6220!6220pairNs#   
    2] a!6220!6220incr#(cons(X, Y)) =#> mark#(X)   
    3] mark#(pairNs) =#> a!6220!6220pairNs#   
    4] mark#(incr(X)) =#> a!6220!6220incr#(mark(X))   
    5] mark#(incr(X)) =#> mark#(X)   
    6] mark#(oddNs) =#> a!6220!6220oddNs#   
    7] mark#(cons(X, Y)) =#> mark#(X)   
    8] mark#(s(X)) =#> mark#(X)   

  Rules R_0:

    a!6220!6220pairNs => cons(0, incr(oddNs)) 
    a!6220!6220oddNs => a!6220!6220incr(a!6220!6220pairNs) 
    a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) 
    mark(pairNs) => a!6220!6220pairNs 
    mark(incr(X)) => a!6220!6220incr(mark(X)) 
    mark(oddNs) => a!6220!6220oddNs 
    mark(cons(X, Y)) => cons(mark(X), Y) 
    mark(0) => 0 
    mark(s(X)) => s(mark(X)) 
    a!6220!6220pairNs => pairNs 
    a!6220!6220incr(X) => incr(X) 
    a!6220!6220oddNs => oddNs 

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 : 2 
    * 1 :  
    * 2 : 3, 4, 5, 6, 7, 8 
    * 3 :  
    * 4 : 2 
    * 5 : 3, 4, 5, 6, 7, 8 
    * 6 : 0, 1 
    * 7 : 3, 4, 5, 6, 7, 8 
    * 8 : 3, 4, 5, 6, 7, 8 

This graph has the following strongly connected components:

  P_1:

    a!6220!6220oddNs# =#> a!6220!6220incr#(a!6220!6220pairNs)   
    a!6220!6220incr#(cons(X, Y)) =#> mark#(X)   
    mark#(incr(X)) =#> a!6220!6220incr#(mark(X))   
    mark#(incr(X)) =#> mark#(X)   
    mark#(oddNs) =#> a!6220!6220oddNs#   
    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!6220pairNs => cons(0, incr(oddNs)) 
  a!6220!6220oddNs => a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) 
  mark(pairNs) => a!6220!6220pairNs 
  mark(incr(X)) => a!6220!6220incr(mark(X)) 
  mark(oddNs) => a!6220!6220oddNs 
  mark(cons(X, Y)) => cons(mark(X), Y) 
  mark(s(X)) => s(mark(X)) 
  a!6220!6220pairNs => pairNs 
  a!6220!6220incr(X) => incr(X) 
  a!6220!6220oddNs => oddNs 

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!6220oddNs# >? a!6220!6220incr#(a!6220!6220pairNs) 
  a!6220!6220incr#(cons(X, Y)) >? mark#(X) 
  mark#(incr(X)) >? a!6220!6220incr#(mark(X)) 
  mark#(incr(X)) >? mark#(X) 
  mark#(oddNs) >? a!6220!6220oddNs# 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  a!6220!6220pairNs >= cons(0, incr(oddNs)) 
  a!6220!6220oddNs >= a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >= cons(s(mark(X)), incr(Y)) 
  mark(pairNs) >= a!6220!6220pairNs 
  mark(incr(X)) >= a!6220!6220incr(mark(X)) 
  mark(oddNs) >= a!6220!6220oddNs 
  mark(cons(X, Y)) >= cons(mark(X), Y) 
  mark(s(X)) >= s(mark(X)) 
  a!6220!6220pairNs >= pairNs 
  a!6220!6220incr(X) >= incr(X) 
  a!6220!6220oddNs >= oddNs 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2y0 
  a!6220!6220incr# = \y0.2y0 
  a!6220!6220oddNs = 1 
  a!6220!6220oddNs# = 0 
  a!6220!6220pairNs = 0 
  cons = \y0y1.2y0 
  incr = \y0.2y0 
  mark = \y0.y0 
  mark# = \y0.2y0 
  oddNs = 1 
  pairNs = 0 
  s = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220oddNs#]] = 0 >= 0 = [[a!6220!6220incr#(a!6220!6220pairNs)]] 
  [[a!6220!6220incr#(cons(_x0, _x1))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 4x0 >= 2x0 = [[a!6220!6220incr#(mark(_x0))]] 
  [[mark#(incr(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(oddNs)]] = 2 > 0 = [[a!6220!6220oddNs#]] 
  [[mark#(cons(_x0, _x1))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 1 >= 0 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 4x0 >= 4x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 1 >= 1 = [[a!6220!6220oddNs]] 
  [[mark(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2x0 >= 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 1 >= 1 = [[oddNs]] 

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!6220oddNs# =#> a!6220!6220incr#(a!6220!6220pairNs)   
  a!6220!6220incr#(cons(X, Y)) =#> mark#(X)   
  mark#(incr(X)) =#> a!6220!6220incr#(mark(X))   
  mark#(incr(X)) =#> mark#(X)   
  mark#(cons(X, Y)) =#> 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 
    * 3 : 2, 3, 4, 5 
    * 4 : 2, 3, 4, 5 
    * 5 : 2, 3, 4, 5 

This graph has the following strongly connected components:

  P_3:

    a!6220!6220incr#(cons(X, Y)) =#> mark#(X)   
    mark#(incr(X)) =#> a!6220!6220incr#(mark(X))   
    mark#(incr(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_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 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) 
  mark#(incr(X)) >? a!6220!6220incr#(mark(X)) 
  mark#(incr(X)) >? mark#(X) 
  mark#(cons(X, Y)) >? mark#(X) 
  mark#(s(X)) >? mark#(X) 
  a!6220!6220pairNs >= cons(0, incr(oddNs)) 
  a!6220!6220oddNs >= a!6220!6220incr(a!6220!6220pairNs) 
  a!6220!6220incr(cons(X, Y)) >= cons(s(mark(X)), incr(Y)) 
  mark(pairNs) >= a!6220!6220pairNs 
  mark(incr(X)) >= a!6220!6220incr(mark(X)) 
  mark(oddNs) >= a!6220!6220oddNs 
  mark(cons(X, Y)) >= cons(mark(X), Y) 
  mark(s(X)) >= s(mark(X)) 
  a!6220!6220pairNs >= pairNs 
  a!6220!6220incr(X) >= incr(X) 
  a!6220!6220oddNs >= oddNs 

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

The following interpretation satisfies the requirements:

  0 = 0 
  a!6220!6220incr = \y0.2 + 2y0 
  a!6220!6220incr# = \y0.2y0 
  a!6220!6220oddNs = 2 
  a!6220!6220pairNs = 0 
  cons = \y0y1.y0 
  incr = \y0.2 + 2y0 
  mark = \y0.2y0 
  mark# = \y0.2y0 
  oddNs = 2 
  pairNs = 0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220incr#(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(incr(_x0))]] = 4 + 4x0 > 4x0 = [[a!6220!6220incr#(mark(_x0))]] 
  [[mark#(incr(_x0))]] = 4 + 4x0 > 2x0 = [[mark#(_x0)]] 
  [[mark#(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[mark#(_x0)]] 
  [[mark#(s(_x0))]] = 2 + 2x0 > 2x0 = [[mark#(_x0)]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[cons(0, incr(oddNs))]] 
  [[a!6220!6220oddNs]] = 2 >= 2 = [[a!6220!6220incr(a!6220!6220pairNs)]] 
  [[a!6220!6220incr(cons(_x0, _x1))]] = 2 + 2x0 >= 1 + 2x0 = [[cons(s(mark(_x0)), incr(_x1))]] 
  [[mark(pairNs)]] = 0 >= 0 = [[a!6220!6220pairNs]] 
  [[mark(incr(_x0))]] = 4 + 4x0 >= 2 + 4x0 = [[a!6220!6220incr(mark(_x0))]] 
  [[mark(oddNs)]] = 4 >= 2 = [[a!6220!6220oddNs]] 
  [[mark(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(mark(_x0), _x1)]] 
  [[mark(s(_x0))]] = 2 + 2x0 >= 1 + 2x0 = [[s(mark(_x0))]] 
  [[a!6220!6220pairNs]] = 0 >= 0 = [[pairNs]] 
  [[a!6220!6220incr(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[incr(_x0)]] 
  [[a!6220!6220oddNs]] = 2 >= 2 = [[oddNs]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, 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 : 1 

This graph has the following strongly connected components:

  P_5:

    mark#(cons(X, Y)) =#> mark#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_4, R_1, m, f) by (P_5, R_1, m, f).

Thus, the original system is terminating if (P_5, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_5, 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#(cons(X, Y))) = cons(X, Y) |> X = nu(mark#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, 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.