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

  !6220!6220 : [o * o] --> o 
  active : [o] --> o 
  and : [o * o] --> o 
  isNePal : [o] --> o 
  mark : [o] --> o 
  nil : [] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  top : [o] --> o 
  tt : [] --> o 

  active(!6220!6220(!6220!6220(X, Y), Z)) => mark(!6220!6220(X, !6220!6220(Y, Z))) 
  active(!6220!6220(X, nil)) => mark(X) 
  active(!6220!6220(nil, X)) => mark(X) 
  active(and(tt, X)) => mark(X) 
  active(isNePal(!6220!6220(X, !6220!6220(Y, X)))) => mark(tt) 
  active(!6220!6220(X, Y)) => !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) => !6220!6220(X, active(Y)) 
  active(and(X, Y)) => and(active(X), Y) 
  active(isNePal(X)) => isNePal(active(X)) 
  !6220!6220(mark(X), Y) => mark(!6220!6220(X, Y)) 
  !6220!6220(X, mark(Y)) => mark(!6220!6220(X, Y)) 
  and(mark(X), Y) => mark(and(X, Y)) 
  isNePal(mark(X)) => mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) => !6220!6220(proper(X), proper(Y)) 
  proper(nil) => ok(nil) 
  proper(and(X, Y)) => and(proper(X), proper(Y)) 
  proper(tt) => ok(tt) 
  proper(isNePal(X)) => isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) => ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) => ok(and(X, Y)) 
  isNePal(ok(X)) => ok(isNePal(X)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(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]):

  active(!6220!6220(!6220!6220(X, Y), Z)) >? mark(!6220!6220(X, !6220!6220(Y, Z))) 
  active(!6220!6220(X, nil)) >? mark(X) 
  active(!6220!6220(nil, X)) >? mark(X) 
  active(and(tt, X)) >? mark(X) 
  active(isNePal(!6220!6220(X, !6220!6220(Y, X)))) >? mark(tt) 
  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(mark(X), Y) >? mark(!6220!6220(X, Y)) 
  !6220!6220(X, mark(Y)) >? mark(!6220!6220(X, Y)) 
  and(mark(X), Y) >? mark(and(X, Y)) 
  isNePal(mark(X)) >? mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(nil) >? ok(nil) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y0 + y1 
  active = \y0.y0 
  and = \y0y1.y0 + y1 
  isNePal = \y0.y0 
  mark = \y0.y0 
  nil = 1 
  ok = \y0.y0 
  proper = \y0.y0 
  top = \y0.2y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(!6220!6220(_x0, _x1), _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[mark(!6220!6220(_x0, !6220!6220(_x1, _x2)))]] 
  [[active(!6220!6220(_x0, nil))]] = 1 + x0 > x0 = [[mark(_x0)]] 
  [[active(!6220!6220(nil, _x0))]] = 1 + x0 > x0 = [[mark(_x0)]] 
  [[active(and(tt, _x0))]] = x0 >= x0 = [[mark(_x0)]] 
  [[active(isNePal(!6220!6220(_x0, !6220!6220(_x1, _x0))))]] = x1 + 2x0 >= 0 = [[mark(tt)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = x0 >= x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(!6220!6220(_x0, _x1))]] 
  [[!6220!6220(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(!6220!6220(_x0, _x1))]] 
  [[and(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(and(_x0, _x1))]] 
  [[isNePal(mark(_x0))]] = x0 >= x0 = [[mark(isNePal(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 1 >= 1 = [[ok(nil)]] 
  [[proper(and(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = x0 >= x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = x0 >= x0 = [[ok(isNePal(_x0))]] 
  [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  active(!6220!6220(X, nil)) => mark(X) 
  active(!6220!6220(nil, X)) => 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]):

  active(!6220!6220(!6220!6220(X, Y), Z)) >? mark(!6220!6220(X, !6220!6220(Y, Z))) 
  active(and(tt, X)) >? mark(X) 
  active(isNePal(!6220!6220(X, !6220!6220(Y, X)))) >? mark(tt) 
  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(mark(X), Y) >? mark(!6220!6220(X, Y)) 
  !6220!6220(X, mark(Y)) >? mark(!6220!6220(X, Y)) 
  and(mark(X), Y) >? mark(and(X, Y)) 
  isNePal(mark(X)) >? mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(nil) >? ok(nil) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y0 + y1 
  active = \y0.y0 
  and = \y0y1.1 + y1 + 2y0 
  isNePal = \y0.2y0 
  mark = \y0.y0 
  nil = 0 
  ok = \y0.y0 
  proper = \y0.y0 
  top = \y0.2y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(!6220!6220(_x0, _x1), _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[mark(!6220!6220(_x0, !6220!6220(_x1, _x2)))]] 
  [[active(and(tt, _x0))]] = 1 + x0 > x0 = [[mark(_x0)]] 
  [[active(isNePal(!6220!6220(_x0, !6220!6220(_x1, _x0))))]] = 2x1 + 4x0 >= 0 = [[mark(tt)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(!6220!6220(_x0, _x1))]] 
  [[!6220!6220(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(!6220!6220(_x0, _x1))]] 
  [[and(mark(_x0), _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(and(_x0, _x1))]] 
  [[isNePal(mark(_x0))]] = 2x0 >= 2x0 = [[mark(isNePal(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] 
  [[proper(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 2x0 >= 2x0 = [[ok(isNePal(_x0))]] 
  [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  active(and(tt, X)) => 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]):

  active(!6220!6220(!6220!6220(X, Y), Z)) >? mark(!6220!6220(X, !6220!6220(Y, Z))) 
  active(isNePal(!6220!6220(X, !6220!6220(Y, X)))) >? mark(tt) 
  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(mark(X), Y) >? mark(!6220!6220(X, Y)) 
  !6220!6220(X, mark(Y)) >? mark(!6220!6220(X, Y)) 
  and(mark(X), Y) >? mark(and(X, Y)) 
  isNePal(mark(X)) >? mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(nil) >? ok(nil) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y1 + 2y0 
  active = \y0.y0 
  and = \y0y1.y1 + 2y0 
  isNePal = \y0.1 + y0 
  mark = \y0.y0 
  nil = 0 
  ok = \y0.y0 
  proper = \y0.y0 
  top = \y0.y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(!6220!6220(_x0, _x1), _x2))]] = x2 + 2x1 + 4x0 >= x2 + 2x0 + 2x1 = [[mark(!6220!6220(_x0, !6220!6220(_x1, _x2)))]] 
  [[active(isNePal(!6220!6220(_x0, !6220!6220(_x1, _x0))))]] = 1 + 2x1 + 3x0 > 0 = [[mark(tt)]] 
  [[active(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 1 + x0 >= 1 + x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(!6220!6220(_x0, _x1))]] 
  [[!6220!6220(_x0, mark(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[mark(!6220!6220(_x0, _x1))]] 
  [[and(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(and(_x0, _x1))]] 
  [[isNePal(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(isNePal(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] 
  [[proper(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = 1 + x0 >= 1 + x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 1 + x0 >= 1 + x0 = [[ok(isNePal(_x0))]] 
  [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  active(isNePal(!6220!6220(X, !6220!6220(Y, X)))) => mark(tt) 

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

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

  active(!6220!6220(!6220!6220(X, Y), Z)) >? mark(!6220!6220(X, !6220!6220(Y, Z))) 
  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(mark(X), Y) >? mark(!6220!6220(X, Y)) 
  !6220!6220(X, mark(Y)) >? mark(!6220!6220(X, Y)) 
  and(mark(X), Y) >? mark(and(X, Y)) 
  isNePal(mark(X)) >? mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(nil) >? ok(nil) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.1 + y1 + 3y0 
  active = \y0.y0 
  and = \y0y1.y1 + 2y0 
  isNePal = \y0.2y0 
  mark = \y0.y0 
  nil = 0 
  ok = \y0.y0 
  proper = \y0.y0 
  top = \y0.y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(!6220!6220(_x0, _x1), _x2))]] = 4 + x2 + 3x1 + 9x0 > 2 + x2 + 3x0 + 3x1 = [[mark(!6220!6220(_x0, !6220!6220(_x1, _x2)))]] 
  [[active(!6220!6220(_x0, _x1))]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(mark(_x0), _x1)]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[mark(!6220!6220(_x0, _x1))]] 
  [[!6220!6220(_x0, mark(_x1))]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[mark(!6220!6220(_x0, _x1))]] 
  [[and(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(and(_x0, _x1))]] 
  [[isNePal(mark(_x0))]] = 2x0 >= 2x0 = [[mark(isNePal(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] 
  [[proper(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 2x0 >= 2x0 = [[ok(isNePal(_x0))]] 
  [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  active(!6220!6220(!6220!6220(X, Y), Z)) => mark(!6220!6220(X, !6220!6220(Y, Z))) 

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

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

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(mark(X), Y) >? mark(!6220!6220(X, Y)) 
  !6220!6220(X, mark(Y)) >? mark(!6220!6220(X, Y)) 
  and(mark(X), Y) >? mark(and(X, Y)) 
  isNePal(mark(X)) >? mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(nil) >? ok(nil) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y1 + 2y0 
  active = \y0.y0 
  and = \y0y1.y1 + 2y0 
  isNePal = \y0.y0 
  mark = \y0.2 + y0 
  nil = 0 
  ok = \y0.y0 
  proper = \y0.y0 
  top = \y0.y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = x0 >= x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(mark(_x0), _x1)]] = 4 + x1 + 2x0 > 2 + x1 + 2x0 = [[mark(!6220!6220(_x0, _x1))]] 
  [[!6220!6220(_x0, mark(_x1))]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[mark(!6220!6220(_x0, _x1))]] 
  [[and(mark(_x0), _x1)]] = 4 + x1 + 2x0 > 2 + x1 + 2x0 = [[mark(and(_x0, _x1))]] 
  [[isNePal(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(isNePal(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] 
  [[proper(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = x0 >= x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = x0 >= x0 = [[ok(isNePal(_x0))]] 
  [[top(mark(_x0))]] = 2 + x0 > x0 = [[top(proper(_x0))]] 
  [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  !6220!6220(mark(X), Y) => mark(!6220!6220(X, Y)) 
  and(mark(X), Y) => mark(and(X, Y)) 
  top(mark(X)) => top(proper(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]):

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(X, mark(Y)) >? mark(!6220!6220(X, Y)) 
  isNePal(mark(X)) >? mark(isNePal(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(nil) >? ok(nil) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y0 + 2y1 
  active = \y0.y0 
  and = \y0y1.y0 + y1 
  isNePal = \y0.2y0 
  mark = \y0.2 + y0 
  nil = 3 
  ok = \y0.2y0 
  proper = \y0.3y0 
  top = \y0.y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(_x0, mark(_x1))]] = 4 + x0 + 2x1 > 2 + x0 + 2x1 = [[mark(!6220!6220(_x0, _x1))]] 
  [[isNePal(mark(_x0))]] = 4 + 2x0 > 2 + 2x0 = [[mark(isNePal(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = 3x0 + 6x1 >= 3x0 + 6x1 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(nil)]] = 9 > 6 = [[ok(nil)]] 
  [[proper(and(_x0, _x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = 6x0 >= 6x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 4x0 >= 4x0 = [[ok(isNePal(_x0))]] 
  [[top(ok(_x0))]] = 2x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  !6220!6220(X, mark(Y)) => mark(!6220!6220(X, Y)) 
  isNePal(mark(X)) => mark(isNePal(X)) 
  proper(nil) => ok(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]):

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  proper(isNePal(X)) >? isNePal(proper(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y0 + 2y1 
  active = \y0.y0 
  and = \y0y1.y0 + 2y1 
  isNePal = \y0.1 + 2y0 
  ok = \y0.y0 
  proper = \y0.2y0 
  top = \y0.y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[isNePal(active(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(and(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[proper(isNePal(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[isNePal(proper(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(isNePal(_x0))]] 
  [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  proper(isNePal(X)) => isNePal(proper(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]):

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  proper(!6220!6220(X, Y)) >? !6220!6220(proper(X), proper(Y)) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.1 + y1 + 2y0 
  active = \y0.y0 
  and = \y0y1.y0 + y1 
  isNePal = \y0.2y0 
  ok = \y0.y0 
  proper = \y0.2y0 
  top = \y0.2y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[proper(!6220!6220(_x0, _x1))]] = 2 + 2x1 + 4x0 > 1 + 2x1 + 4x0 = [[!6220!6220(proper(_x0), proper(_x1))]] 
  [[proper(and(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 2x0 >= 2x0 = [[ok(isNePal(_x0))]] 
  [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  proper(!6220!6220(X, Y)) => !6220!6220(proper(X), proper(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]):

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  proper(tt) >? ok(tt) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y0 + y1 
  active = \y0.y0 
  and = \y0y1.y0 + 2y1 
  isNePal = \y0.2y0 
  ok = \y0.2y0 
  proper = \y0.3y0 
  top = \y0.2y0 
  tt = 2 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[proper(and(_x0, _x1))]] = 3x0 + 6x1 >= 3x0 + 6x1 = [[and(proper(_x0), proper(_x1))]] 
  [[proper(tt)]] = 6 > 4 = [[ok(tt)]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 4x0 >= 4x0 = [[ok(isNePal(_x0))]] 
  [[top(ok(_x0))]] = 4x0 >= 2x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  proper(tt) => ok(tt) 

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

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

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  proper(and(X, Y)) >? and(proper(X), proper(Y)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y1 + 2y0 
  active = \y0.y0 
  and = \y0y1.2 + y1 + 2y0 
  isNePal = \y0.2y0 
  ok = \y0.y0 
  proper = \y0.2y0 
  top = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[proper(and(_x0, _x1))]] = 4 + 2x1 + 4x0 > 2 + 2x1 + 4x0 = [[and(proper(_x0), proper(_x1))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 2x0 >= 2x0 = [[ok(isNePal(_x0))]] 
  [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  proper(and(X, Y)) => and(proper(X), proper(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]):

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 
  !6220!6220(ok(X), ok(Y)) >? ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) >? ok(and(X, Y)) 
  isNePal(ok(X)) >? ok(isNePal(X)) 
  top(ok(X)) >? top(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.y0 + y1 
  active = \y0.y0 
  and = \y0y1.y1 + 2y0 
  isNePal = \y0.2y0 
  ok = \y0.1 + 2y0 
  top = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 2x0 >= 2x0 = [[isNePal(active(_x0))]] 
  [[!6220!6220(ok(_x0), ok(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[ok(!6220!6220(_x0, _x1))]] 
  [[and(ok(_x0), ok(_x1))]] = 3 + 2x1 + 4x0 > 1 + 2x1 + 4x0 = [[ok(and(_x0, _x1))]] 
  [[isNePal(ok(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[ok(isNePal(_x0))]] 
  [[top(ok(_x0))]] = 1 + 2x0 > x0 = [[top(active(_x0))]] 

We can thus remove the following rules:

  !6220!6220(ok(X), ok(Y)) => ok(!6220!6220(X, Y)) 
  and(ok(X), ok(Y)) => ok(and(X, Y)) 
  isNePal(ok(X)) => ok(isNePal(X)) 
  top(ok(X)) => top(active(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]):

  active(!6220!6220(X, Y)) >? !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) >? !6220!6220(X, active(Y)) 
  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 

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

The following interpretation satisfies the requirements:

  !6220!6220 = \y0y1.3 + y0 + y1 
  active = \y0.3y0 
  and = \y0y1.y0 + y1 
  isNePal = \y0.y0 

Using this interpretation, the requirements translate to:

  [[active(!6220!6220(_x0, _x1))]] = 9 + 3x0 + 3x1 > 3 + x1 + 3x0 = [[!6220!6220(active(_x0), _x1)]] 
  [[active(!6220!6220(_x0, _x1))]] = 9 + 3x0 + 3x1 > 3 + x0 + 3x1 = [[!6220!6220(_x0, active(_x1))]] 
  [[active(and(_x0, _x1))]] = 3x0 + 3x1 >= x1 + 3x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 3x0 >= 3x0 = [[isNePal(active(_x0))]] 

We can thus remove the following rules:

  active(!6220!6220(X, Y)) => !6220!6220(active(X), Y) 
  active(!6220!6220(X, Y)) => !6220!6220(X, active(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]):

  active(and(X, Y)) >? and(active(X), Y) 
  active(isNePal(X)) >? isNePal(active(X)) 

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

The following interpretation satisfies the requirements:

  active = \y0.3y0 
  and = \y0y1.y0 + y1 
  isNePal = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[active(and(_x0, _x1))]] = 3x0 + 3x1 >= x1 + 3x0 = [[and(active(_x0), _x1)]] 
  [[active(isNePal(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[isNePal(active(_x0))]] 

We can thus remove the following rules:

  active(isNePal(X)) => isNePal(active(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]):

  active(and(X, Y)) >? and(active(X), Y) 

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

The following interpretation satisfies the requirements:

  active = \y0.3y0 
  and = \y0y1.1 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[active(and(_x0, _x1))]] = 3 + 3x0 + 3x1 > 1 + x1 + 3x0 = [[and(active(_x0), _x1)]] 

We can thus remove the following rules:

  active(and(X, Y)) => and(active(X), Y) 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.