/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.

  !minus : [o * o] --> o 
  !plus : [o * o] --> o 
  0 : [] --> o 
  1 : [] --> o 
  BS : [o] --> o 
  I : [o] --> o 
  L : [o] --> o 
  Log : [o] --> o 
  Log!450 : [o] --> o 
  Max : [o] --> o 
  Min : [o] --> o 
  N : [o * o * o] --> o 
  O : [o] --> o 
  Size : [o] --> o 
  Val : [o] --> o 
  WB : [o] --> o 
  and : [o * o] --> o 
  false : [] --> o 
  ge : [o * o] --> o 
  if : [o * o * o] --> o 
  l : [] --> o 
  not : [o] --> o 
  r : [] --> o 
  true : [] --> o 

  O(0) => 0 
  !plus(0, X) => X 
  !plus(X, 0) => X 
  !plus(O(X), O(Y)) => O(!plus(X, Y)) 
  !plus(O(X), I(Y)) => I(!plus(X, Y)) 
  !plus(I(X), O(Y)) => I(!plus(X, Y)) 
  !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 
  !minus(X, 0) => X 
  !minus(0, X) => 0 
  !minus(O(X), O(Y)) => O(!minus(X, Y)) 
  !minus(O(X), I(Y)) => I(!minus(!minus(X, Y), I(1))) 
  !minus(I(X), O(Y)) => I(!minus(X, Y)) 
  !minus(I(X), I(Y)) => O(!minus(X, Y)) 
  not(true) => false 
  not(false) => true 
  and(X, true) => X 
  and(X, false) => false 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  ge(O(X), O(Y)) => ge(X, Y) 
  ge(O(X), I(Y)) => not(ge(Y, X)) 
  ge(I(X), O(Y)) => ge(X, Y) 
  ge(I(X), I(Y)) => ge(X, Y) 
  ge(X, 0) => true 
  ge(0, O(X)) => ge(0, X) 
  ge(0, I(X)) => false 
  Log!450(0) => 0 
  Log!450(I(X)) => !plus(Log!450(X), I(0)) 
  Log!450(O(X)) => if(ge(X, I(0)), !plus(Log!450(X), I(0)), 0) 
  Log(X) => !minus(Log!450(X), I(0)) 
  Val(L(X)) => X 
  Val(N(X, l, r)) => X 
  Min(L(X)) => X 
  Min(N(X, l, r)) => Min(l) 
  Max(L(X)) => X 
  Max(N(X, l, r)) => Max(r) 
  BS(L(X)) => true 
  BS(N(X, l, r)) => and(and(ge(X, Max(l)), ge(Min(r), X)), and(BS(l), BS(r))) 
  Size(L(X)) => I(0) 
  Size(N(X, l, r)) => !plus(!plus(Size(l), Size(r)), I(1)) 
  WB(L(X)) => true 
  WB(N(X, l, r)) => and(if(ge(Size(l), Size(r)), ge(I(0), !minus(Size(l), Size(r))), ge(I(0), !minus(Size(r), Size(l)))), and(WB(l), WB(r))) 

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

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

  O(0) >? 0 
  !plus(0, X) >? X 
  !plus(X, 0) >? X 
  !plus(O(X), O(Y)) >? O(!plus(X, Y)) 
  !plus(O(X), I(Y)) >? I(!plus(X, Y)) 
  !plus(I(X), O(Y)) >? I(!plus(X, Y)) 
  !plus(I(X), I(Y)) >? O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 
  !minus(X, 0) >? X 
  !minus(0, X) >? 0 
  !minus(O(X), O(Y)) >? O(!minus(X, Y)) 
  !minus(O(X), I(Y)) >? I(!minus(!minus(X, Y), I(1))) 
  !minus(I(X), O(Y)) >? I(!minus(X, Y)) 
  !minus(I(X), I(Y)) >? O(!minus(X, Y)) 
  not(true) >? false 
  not(false) >? true 
  and(X, true) >? X 
  and(X, false) >? false 
  if(true, X, Y) >? X 
  if(false, X, Y) >? Y 
  ge(O(X), O(Y)) >? ge(X, Y) 
  ge(O(X), I(Y)) >? not(ge(Y, X)) 
  ge(I(X), O(Y)) >? ge(X, Y) 
  ge(I(X), I(Y)) >? ge(X, Y) 
  ge(X, 0) >? true 
  ge(0, O(X)) >? ge(0, X) 
  ge(0, I(X)) >? false 
  Log!450(0) >? 0 
  Log!450(I(X)) >? !plus(Log!450(X), I(0)) 
  Log!450(O(X)) >? if(ge(X, I(0)), !plus(Log!450(X), I(0)), 0) 
  Log(X) >? !minus(Log!450(X), I(0)) 
  Val(L(X)) >? X 
  Val(N(X, l, r)) >? X 
  Min(L(X)) >? X 
  Min(N(X, l, r)) >? Min(l) 
  Max(L(X)) >? X 
  Max(N(X, l, r)) >? Max(r) 
  BS(L(X)) >? true 
  BS(N(X, l, r)) >? and(and(ge(X, Max(l)), ge(Min(r), X)), and(BS(l), BS(r))) 
  Size(L(X)) >? I(0) 
  Size(N(X, l, r)) >? !plus(!plus(Size(l), Size(r)), I(1)) 
  WB(L(X)) >? true 
  WB(N(X, l, r)) >? and(if(ge(Size(l), Size(r)), ge(I(0), !minus(Size(l), Size(r))), ge(I(0), !minus(Size(r), Size(l)))), and(WB(l), WB(r))) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 + y1 
  !plus = \y0y1.y0 + 2y1 
  0 = 0 
  1 = 0 
  BS = \y0.2y0 
  I = \y0.2y0 
  L = \y0.3 + y0 
  Log = \y0.3 + 3y0 
  Log!450 = \y0.2y0 
  Max = \y0.2y0 
  Min = \y0.2y0 
  N = \y0y1y2.3 + 3y0 + 3y1 + 3y2 
  O = \y0.2y0 
  Size = \y0.y0 
  Val = \y0.3 + 2y0 
  WB = \y0.y0 
  and = \y0y1.y0 + 2y1 
  false = 0 
  ge = \y0y1.2y0 + 2y1 
  if = \y0y1y2.y0 + y1 + y2 
  l = 0 
  not = \y0.2y0 
  r = 0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[O(0)]] = 0 >= 0 = [[0]] 
  [[!plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(O(_x0), O(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[O(!plus(_x0, _x1))]] 
  [[!plus(O(_x0), I(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), O(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), I(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[O(!plus(!plus(_x0, _x1), I(0)))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = x0 + 2x1 + 4x2 >= x0 + 2x1 + 2x2 = [[!plus(!plus(_x0, _x1), _x2)]] 
  [[!minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!minus(0, _x0)]] = x0 >= 0 = [[0]] 
  [[!minus(O(_x0), O(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[O(!minus(_x0, _x1))]] 
  [[!minus(O(_x0), I(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[I(!minus(!minus(_x0, _x1), I(1)))]] 
  [[!minus(I(_x0), O(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[I(!minus(_x0, _x1))]] 
  [[!minus(I(_x0), I(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[O(!minus(_x0, _x1))]] 
  [[not(true)]] = 0 >= 0 = [[false]] 
  [[not(false)]] = 0 >= 0 = [[true]] 
  [[and(_x0, true)]] = x0 >= x0 = [[_x0]] 
  [[and(_x0, false)]] = x0 >= 0 = [[false]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[ge(O(_x0), O(_x1))]] = 4x0 + 4x1 >= 2x0 + 2x1 = [[ge(_x0, _x1)]] 
  [[ge(O(_x0), I(_x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[not(ge(_x1, _x0))]] 
  [[ge(I(_x0), O(_x1))]] = 4x0 + 4x1 >= 2x0 + 2x1 = [[ge(_x0, _x1)]] 
  [[ge(I(_x0), I(_x1))]] = 4x0 + 4x1 >= 2x0 + 2x1 = [[ge(_x0, _x1)]] 
  [[ge(_x0, 0)]] = 2x0 >= 0 = [[true]] 
  [[ge(0, O(_x0))]] = 4x0 >= 2x0 = [[ge(0, _x0)]] 
  [[ge(0, I(_x0))]] = 4x0 >= 0 = [[false]] 
  [[Log!450(0)]] = 0 >= 0 = [[0]] 
  [[Log!450(I(_x0))]] = 4x0 >= 2x0 = [[!plus(Log!450(_x0), I(0))]] 
  [[Log!450(O(_x0))]] = 4x0 >= 4x0 = [[if(ge(_x0, I(0)), !plus(Log!450(_x0), I(0)), 0)]] 
  [[Log(_x0)]] = 3 + 3x0 > 2x0 = [[!minus(Log!450(_x0), I(0))]] 
  [[Val(L(_x0))]] = 9 + 2x0 > x0 = [[_x0]] 
  [[Val(N(_x0, l, r))]] = 9 + 6x0 > x0 = [[_x0]] 
  [[Min(L(_x0))]] = 6 + 2x0 > x0 = [[_x0]] 
  [[Min(N(_x0, l, r))]] = 6 + 6x0 > 0 = [[Min(l)]] 
  [[Max(L(_x0))]] = 6 + 2x0 > x0 = [[_x0]] 
  [[Max(N(_x0, l, r))]] = 6 + 6x0 > 0 = [[Max(r)]] 
  [[BS(L(_x0))]] = 6 + 2x0 > 0 = [[true]] 
  [[BS(N(_x0, l, r))]] = 6 + 6x0 > 6x0 = [[and(and(ge(_x0, Max(l)), ge(Min(r), _x0)), and(BS(l), BS(r)))]] 
  [[Size(L(_x0))]] = 3 + x0 > 0 = [[I(0)]] 
  [[Size(N(_x0, l, r))]] = 3 + 3x0 > 0 = [[!plus(!plus(Size(l), Size(r)), I(1))]] 
  [[WB(L(_x0))]] = 3 + x0 > 0 = [[true]] 
  [[WB(N(_x0, l, r))]] = 3 + 3x0 > 0 = [[and(if(ge(Size(l), Size(r)), ge(I(0), !minus(Size(l), Size(r))), ge(I(0), !minus(Size(r), Size(l)))), and(WB(l), WB(r)))]] 

We can thus remove the following rules:

  Log(X) => !minus(Log!450(X), I(0)) 
  Val(L(X)) => X 
  Val(N(X, l, r)) => X 
  Min(L(X)) => X 
  Min(N(X, l, r)) => Min(l) 
  Max(L(X)) => X 
  Max(N(X, l, r)) => Max(r) 
  BS(L(X)) => true 
  BS(N(X, l, r)) => and(and(ge(X, Max(l)), ge(Min(r), X)), and(BS(l), BS(r))) 
  Size(L(X)) => I(0) 
  Size(N(X, l, r)) => !plus(!plus(Size(l), Size(r)), I(1)) 
  WB(L(X)) => true 
  WB(N(X, l, r)) => and(if(ge(Size(l), Size(r)), ge(I(0), !minus(Size(l), Size(r))), ge(I(0), !minus(Size(r), Size(l)))), and(WB(l), WB(r))) 

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

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

  O(0) >? 0 
  !plus(0, X) >? X 
  !plus(X, 0) >? X 
  !plus(O(X), O(Y)) >? O(!plus(X, Y)) 
  !plus(O(X), I(Y)) >? I(!plus(X, Y)) 
  !plus(I(X), O(Y)) >? I(!plus(X, Y)) 
  !plus(I(X), I(Y)) >? O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 
  !minus(X, 0) >? X 
  !minus(0, X) >? 0 
  !minus(O(X), O(Y)) >? O(!minus(X, Y)) 
  !minus(O(X), I(Y)) >? I(!minus(!minus(X, Y), I(1))) 
  !minus(I(X), O(Y)) >? I(!minus(X, Y)) 
  !minus(I(X), I(Y)) >? O(!minus(X, Y)) 
  not(true) >? false 
  not(false) >? true 
  and(X, true) >? X 
  and(X, false) >? false 
  if(true, X, Y) >? X 
  if(false, X, Y) >? Y 
  ge(O(X), O(Y)) >? ge(X, Y) 
  ge(O(X), I(Y)) >? not(ge(Y, X)) 
  ge(I(X), O(Y)) >? ge(X, Y) 
  ge(I(X), I(Y)) >? ge(X, Y) 
  ge(X, 0) >? true 
  ge(0, O(X)) >? ge(0, X) 
  ge(0, I(X)) >? false 
  Log!450(0) >? 0 
  Log!450(I(X)) >? !plus(Log!450(X), I(0)) 
  Log!450(O(X)) >? if(ge(X, I(0)), !plus(Log!450(X), I(0)), 0) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 + 2y1 
  !plus = \y0y1.y0 + 2y1 
  0 = 0 
  1 = 0 
  I = \y0.2y0 
  Log!450 = \y0.2y0 
  O = \y0.2y0 
  and = \y0y1.3 + y0 + 3y1 
  false = 0 
  ge = \y0y1.y0 + 2y1 
  if = \y0y1y2.y0 + y1 + y2 
  not = \y0.y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[O(0)]] = 0 >= 0 = [[0]] 
  [[!plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(O(_x0), O(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[O(!plus(_x0, _x1))]] 
  [[!plus(O(_x0), I(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), O(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), I(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[O(!plus(!plus(_x0, _x1), I(0)))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = x0 + 2x1 + 4x2 >= x0 + 2x1 + 2x2 = [[!plus(!plus(_x0, _x1), _x2)]] 
  [[!minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!minus(0, _x0)]] = 2x0 >= 0 = [[0]] 
  [[!minus(O(_x0), O(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[O(!minus(_x0, _x1))]] 
  [[!minus(O(_x0), I(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[I(!minus(!minus(_x0, _x1), I(1)))]] 
  [[!minus(I(_x0), O(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[I(!minus(_x0, _x1))]] 
  [[!minus(I(_x0), I(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[O(!minus(_x0, _x1))]] 
  [[not(true)]] = 0 >= 0 = [[false]] 
  [[not(false)]] = 0 >= 0 = [[true]] 
  [[and(_x0, true)]] = 3 + x0 > x0 = [[_x0]] 
  [[and(_x0, false)]] = 3 + x0 > 0 = [[false]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[ge(O(_x0), O(_x1))]] = 2x0 + 4x1 >= x0 + 2x1 = [[ge(_x0, _x1)]] 
  [[ge(O(_x0), I(_x1))]] = 2x0 + 4x1 >= x1 + 2x0 = [[not(ge(_x1, _x0))]] 
  [[ge(I(_x0), O(_x1))]] = 2x0 + 4x1 >= x0 + 2x1 = [[ge(_x0, _x1)]] 
  [[ge(I(_x0), I(_x1))]] = 2x0 + 4x1 >= x0 + 2x1 = [[ge(_x0, _x1)]] 
  [[ge(_x0, 0)]] = x0 >= 0 = [[true]] 
  [[ge(0, O(_x0))]] = 4x0 >= 2x0 = [[ge(0, _x0)]] 
  [[ge(0, I(_x0))]] = 4x0 >= 0 = [[false]] 
  [[Log!450(0)]] = 0 >= 0 = [[0]] 
  [[Log!450(I(_x0))]] = 4x0 >= 2x0 = [[!plus(Log!450(_x0), I(0))]] 
  [[Log!450(O(_x0))]] = 4x0 >= 3x0 = [[if(ge(_x0, I(0)), !plus(Log!450(_x0), I(0)), 0)]] 

We can thus remove the following rules:

  and(X, true) => X 
  and(X, false) => false 

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

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

  O(0) >? 0 
  !plus(0, X) >? X 
  !plus(X, 0) >? X 
  !plus(O(X), O(Y)) >? O(!plus(X, Y)) 
  !plus(O(X), I(Y)) >? I(!plus(X, Y)) 
  !plus(I(X), O(Y)) >? I(!plus(X, Y)) 
  !plus(I(X), I(Y)) >? O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 
  !minus(X, 0) >? X 
  !minus(0, X) >? 0 
  !minus(O(X), O(Y)) >? O(!minus(X, Y)) 
  !minus(O(X), I(Y)) >? I(!minus(!minus(X, Y), I(1))) 
  !minus(I(X), O(Y)) >? I(!minus(X, Y)) 
  !minus(I(X), I(Y)) >? O(!minus(X, Y)) 
  not(true) >? false 
  not(false) >? true 
  if(true, X, Y) >? X 
  if(false, X, Y) >? Y 
  ge(O(X), O(Y)) >? ge(X, Y) 
  ge(O(X), I(Y)) >? not(ge(Y, X)) 
  ge(I(X), O(Y)) >? ge(X, Y) 
  ge(I(X), I(Y)) >? ge(X, Y) 
  ge(X, 0) >? true 
  ge(0, O(X)) >? ge(0, X) 
  ge(0, I(X)) >? false 
  Log!450(0) >? 0 
  Log!450(I(X)) >? !plus(Log!450(X), I(0)) 
  Log!450(O(X)) >? if(ge(X, I(0)), !plus(Log!450(X), I(0)), 0) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 + 2y1 
  !plus = \y0y1.y0 + y1 
  0 = 0 
  1 = 0 
  I = \y0.3y0 
  Log!450 = \y0.1 + y0 
  O = \y0.3y0 
  false = 0 
  ge = \y0y1.y1 + 2y0 
  if = \y0y1y2.y0 + y1 + y2 
  not = \y0.y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[O(0)]] = 0 >= 0 = [[0]] 
  [[!plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(O(_x0), O(_x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[O(!plus(_x0, _x1))]] 
  [[!plus(O(_x0), I(_x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), O(_x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), I(_x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[O(!plus(!plus(_x0, _x1), I(0)))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(!plus(_x0, _x1), _x2)]] 
  [[!minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!minus(0, _x0)]] = 2x0 >= 0 = [[0]] 
  [[!minus(O(_x0), O(_x1))]] = 3x0 + 6x1 >= 3x0 + 6x1 = [[O(!minus(_x0, _x1))]] 
  [[!minus(O(_x0), I(_x1))]] = 3x0 + 6x1 >= 3x0 + 6x1 = [[I(!minus(!minus(_x0, _x1), I(1)))]] 
  [[!minus(I(_x0), O(_x1))]] = 3x0 + 6x1 >= 3x0 + 6x1 = [[I(!minus(_x0, _x1))]] 
  [[!minus(I(_x0), I(_x1))]] = 3x0 + 6x1 >= 3x0 + 6x1 = [[O(!minus(_x0, _x1))]] 
  [[not(true)]] = 0 >= 0 = [[false]] 
  [[not(false)]] = 0 >= 0 = [[true]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[ge(O(_x0), O(_x1))]] = 3x1 + 6x0 >= x1 + 2x0 = [[ge(_x0, _x1)]] 
  [[ge(O(_x0), I(_x1))]] = 3x1 + 6x0 >= x0 + 2x1 = [[not(ge(_x1, _x0))]] 
  [[ge(I(_x0), O(_x1))]] = 3x1 + 6x0 >= x1 + 2x0 = [[ge(_x0, _x1)]] 
  [[ge(I(_x0), I(_x1))]] = 3x1 + 6x0 >= x1 + 2x0 = [[ge(_x0, _x1)]] 
  [[ge(_x0, 0)]] = 2x0 >= 0 = [[true]] 
  [[ge(0, O(_x0))]] = 3x0 >= x0 = [[ge(0, _x0)]] 
  [[ge(0, I(_x0))]] = 3x0 >= 0 = [[false]] 
  [[Log!450(0)]] = 1 > 0 = [[0]] 
  [[Log!450(I(_x0))]] = 1 + 3x0 >= 1 + x0 = [[!plus(Log!450(_x0), I(0))]] 
  [[Log!450(O(_x0))]] = 1 + 3x0 >= 1 + 3x0 = [[if(ge(_x0, I(0)), !plus(Log!450(_x0), I(0)), 0)]] 

We can thus remove the following rules:

  Log!450(0) => 0 

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] !plus#(O(X), O(Y)) =#> O#(!plus(X, Y))   
    1] !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
    2] !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
    3] !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
    4] !plus#(I(X), I(Y)) =#> O#(!plus(!plus(X, Y), I(0)))   
    5] !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
    6] !plus#(I(X), I(Y)) =#> !plus#(X, Y)   
    7] !plus#(X, !plus(Y, Z)) =#> !plus#(!plus(X, Y), Z)   
    8] !plus#(X, !plus(Y, Z)) =#> !plus#(X, Y)   
    9] !minus#(O(X), O(Y)) =#> O#(!minus(X, Y))   
    10] !minus#(O(X), O(Y)) =#> !minus#(X, Y)   
    11] !minus#(O(X), I(Y)) =#> !minus#(!minus(X, Y), I(1))   
    12] !minus#(O(X), I(Y)) =#> !minus#(X, Y)   
    13] !minus#(I(X), O(Y)) =#> !minus#(X, Y)   
    14] !minus#(I(X), I(Y)) =#> O#(!minus(X, Y))   
    15] !minus#(I(X), I(Y)) =#> !minus#(X, Y)   
    16] ge#(O(X), O(Y)) =#> ge#(X, Y)   
    17] ge#(O(X), I(Y)) =#> not#(ge(Y, X))   
    18] ge#(O(X), I(Y)) =#> ge#(Y, X)   
    19] ge#(I(X), O(Y)) =#> ge#(X, Y)   
    20] ge#(I(X), I(Y)) =#> ge#(X, Y)   
    21] ge#(0, O(X)) =#> ge#(0, X)   
    22] Log!450#(I(X)) =#> !plus#(Log!450(X), I(0))   
    23] Log!450#(I(X)) =#> Log!450#(X)   
    24] Log!450#(O(X)) =#> if#(ge(X, I(0)), !plus(Log!450(X), I(0)), 0)   
    25] Log!450#(O(X)) =#> ge#(X, I(0))   
    26] Log!450#(O(X)) =#> !plus#(Log!450(X), I(0))   
    27] Log!450#(O(X)) =#> Log!450#(X)   

  Rules R_0:

    O(0) => 0 
    !plus(0, X) => X 
    !plus(X, 0) => X 
    !plus(O(X), O(Y)) => O(!plus(X, Y)) 
    !plus(O(X), I(Y)) => I(!plus(X, Y)) 
    !plus(I(X), O(Y)) => I(!plus(X, Y)) 
    !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
    !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 
    !minus(X, 0) => X 
    !minus(0, X) => 0 
    !minus(O(X), O(Y)) => O(!minus(X, Y)) 
    !minus(O(X), I(Y)) => I(!minus(!minus(X, Y), I(1))) 
    !minus(I(X), O(Y)) => I(!minus(X, Y)) 
    !minus(I(X), I(Y)) => O(!minus(X, Y)) 
    not(true) => false 
    not(false) => true 
    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    ge(O(X), O(Y)) => ge(X, Y) 
    ge(O(X), I(Y)) => not(ge(Y, X)) 
    ge(I(X), O(Y)) => ge(X, Y) 
    ge(I(X), I(Y)) => ge(X, Y) 
    ge(X, 0) => true 
    ge(0, O(X)) => ge(0, X) 
    ge(0, I(X)) => false 
    Log!450(I(X)) => !plus(Log!450(X), I(0)) 
    Log!450(O(X)) => if(ge(X, I(0)), !plus(Log!450(X), I(0)), 0) 

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 :  
    * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 4 :  
    * 5 : 2, 4, 5, 6 
    * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8 
    * 9 :  
    * 10 : 9, 10, 11, 12, 13, 14, 15 
    * 11 : 11, 12, 14, 15 
    * 12 : 9, 10, 11, 12, 13, 14, 15 
    * 13 : 9, 10, 11, 12, 13, 14, 15 
    * 14 :  
    * 15 : 9, 10, 11, 12, 13, 14, 15 
    * 16 : 16, 17, 18, 19, 20, 21 
    * 17 :  
    * 18 : 16, 17, 18, 19, 20, 21 
    * 19 : 16, 17, 18, 19, 20, 21 
    * 20 : 16, 17, 18, 19, 20, 21 
    * 21 : 21 
    * 22 : 2, 4, 5, 6 
    * 23 : 22, 23, 24, 25, 26, 27 
    * 24 :  
    * 25 : 17, 18, 20 
    * 26 : 2, 4, 5, 6 
    * 27 : 22, 23, 24, 25, 26, 27 

This graph has the following strongly connected components:

  P_1:

    !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
    !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
    !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
    !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
    !plus#(I(X), I(Y)) =#> !plus#(X, Y)   
    !plus#(X, !plus(Y, Z)) =#> !plus#(!plus(X, Y), Z)   
    !plus#(X, !plus(Y, Z)) =#> !plus#(X, Y)   

  P_2:

    !minus#(O(X), O(Y)) =#> !minus#(X, Y)   
    !minus#(O(X), I(Y)) =#> !minus#(!minus(X, Y), I(1))   
    !minus#(O(X), I(Y)) =#> !minus#(X, Y)   
    !minus#(I(X), O(Y)) =#> !minus#(X, Y)   
    !minus#(I(X), I(Y)) =#> !minus#(X, Y)   

  P_3:

    ge#(O(X), O(Y)) =#> ge#(X, Y)   
    ge#(O(X), I(Y)) =#> ge#(Y, X)   
    ge#(I(X), O(Y)) =#> ge#(X, Y)   
    ge#(I(X), I(Y)) =#> ge#(X, Y)   

  P_4:

    ge#(0, O(X)) =#> ge#(0, X)   

  P_5:

    Log!450#(I(X)) =#> Log!450#(X)   
    Log!450#(O(X)) =#> Log!450#(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), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f) and (P_5, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_5, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(Log!450#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(Log!450#(I(X))) = I(X) |> X = nu(Log!450#(X)) 
  nu(Log!450#(O(X))) = O(X) |> X = nu(Log!450#(X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(ge#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(ge#(0, O(X))) = O(X) |> X = nu(ge#(0, X)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_3, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  ge#(O(X), O(Y)) >? ge#(X, Y) 
  ge#(O(X), I(Y)) >? ge#(Y, X) 
  ge#(I(X), O(Y)) >? ge#(X, Y) 
  ge#(I(X), I(Y)) >? ge#(X, Y) 

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

The following interpretation satisfies the requirements:

  I = \y0.3 + y0 
  O = \y0.3 + y0 
  ge# = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[ge#(O(_x0), O(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x0, _x1)]] 
  [[ge#(O(_x0), I(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x1, _x0)]] 
  [[ge#(I(_x0), O(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x0, _x1)]] 
  [[ge#(I(_x0), I(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_2, R_0) are:

  O(0) => 0 
  !minus(X, 0) => X 
  !minus(0, X) => 0 
  !minus(O(X), O(Y)) => O(!minus(X, Y)) 
  !minus(O(X), I(Y)) => I(!minus(!minus(X, Y), I(1))) 
  !minus(I(X), O(Y)) => I(!minus(X, Y)) 
  !minus(I(X), I(Y)) => O(!minus(X, Y)) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  !minus#(O(X), O(Y)) >? !minus#(X, Y) 
  !minus#(O(X), I(Y)) >? !minus#(!minus(X, Y), I(1)) 
  !minus#(O(X), I(Y)) >? !minus#(X, Y) 
  !minus#(I(X), O(Y)) >? !minus#(X, Y) 
  !minus#(I(X), I(Y)) >? !minus#(X, Y) 
  O(0) >= 0 
  !minus(X, 0) >= X 
  !minus(0, X) >= 0 
  !minus(O(X), O(Y)) >= O(!minus(X, Y)) 
  !minus(O(X), I(Y)) >= I(!minus(!minus(X, Y), I(1))) 
  !minus(I(X), O(Y)) >= I(!minus(X, Y)) 
  !minus(I(X), I(Y)) >= O(!minus(X, Y)) 

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

The following interpretation satisfies the requirements:

  !minus = \y0y1.y0 
  !minus# = \y0y1.y0 
  0 = 1 
  1 = 0 
  I = \y0.1 + y0 
  O = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[!minus#(O(_x0), O(_x1))]] = 1 + x0 > x0 = [[!minus#(_x0, _x1)]] 
  [[!minus#(O(_x0), I(_x1))]] = 1 + x0 > x0 = [[!minus#(!minus(_x0, _x1), I(1))]] 
  [[!minus#(O(_x0), I(_x1))]] = 1 + x0 > x0 = [[!minus#(_x0, _x1)]] 
  [[!minus#(I(_x0), O(_x1))]] = 1 + x0 > x0 = [[!minus#(_x0, _x1)]] 
  [[!minus#(I(_x0), I(_x1))]] = 1 + x0 > x0 = [[!minus#(_x0, _x1)]] 
  [[O(0)]] = 2 >= 1 = [[0]] 
  [[!minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!minus(0, _x0)]] = 1 >= 1 = [[0]] 
  [[!minus(O(_x0), O(_x1))]] = 1 + x0 >= 1 + x0 = [[O(!minus(_x0, _x1))]] 
  [[!minus(O(_x0), I(_x1))]] = 1 + x0 >= 1 + x0 = [[I(!minus(!minus(_x0, _x1), I(1)))]] 
  [[!minus(I(_x0), O(_x1))]] = 1 + x0 >= 1 + x0 = [[I(!minus(_x0, _x1))]] 
  [[!minus(I(_x0), I(_x1))]] = 1 + x0 >= 1 + x0 = [[O(!minus(_x0, _x1))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_0) are:

  O(0) => 0 
  !plus(0, X) => X 
  !plus(X, 0) => X 
  !plus(O(X), O(Y)) => O(!plus(X, Y)) 
  !plus(O(X), I(Y)) => I(!plus(X, Y)) 
  !plus(I(X), O(Y)) => I(!plus(X, Y)) 
  !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  !plus#(O(X), O(Y)) >? !plus#(X, Y) 
  !plus#(O(X), I(Y)) >? !plus#(X, Y) 
  !plus#(I(X), O(Y)) >? !plus#(X, Y) 
  !plus#(I(X), I(Y)) >? !plus#(!plus(X, Y), I(0)) 
  !plus#(I(X), I(Y)) >? !plus#(X, Y) 
  !plus#(X, !plus(Y, Z)) >? !plus#(!plus(X, Y), Z) 
  !plus#(X, !plus(Y, Z)) >? !plus#(X, Y) 
  O(0) >= 0 
  !plus(0, X) >= X 
  !plus(X, 0) >= X 
  !plus(O(X), O(Y)) >= O(!plus(X, Y)) 
  !plus(O(X), I(Y)) >= I(!plus(X, Y)) 
  !plus(I(X), O(Y)) >= I(!plus(X, Y)) 
  !plus(I(X), I(Y)) >= O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) >= !plus(!plus(X, Y), Z) 

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

We consider usable_rules with respect to the following argument filtering:

  !plus#(x_1,x_2) = !plus#(x_2) 

This leaves the following ordering requirements:

  !plus#(O(X), O(Y)) >= !plus#(X, Y) 
  !plus#(O(X), I(Y)) >= !plus#(X, Y) 
  !plus#(I(X), O(Y)) >= !plus#(X, Y) 
  !plus#(I(X), I(Y)) >= !plus#(!plus(X, Y), I(0)) 
  !plus#(I(X), I(Y)) >= !plus#(X, Y) 
  !plus#(X, !plus(Y, Z)) > !plus#(!plus(X, Y), Z) 
  !plus#(X, !plus(Y, Z)) >= !plus#(X, Y) 

The following interpretation satisfies the requirements:

  !plus = \y0y1.2 + y1 + 2y0 
  !plus# = \y0y1.y1 
  0 = 0 
  I = \y0.2y0 
  O = \y0.y0 

Using this interpretation, the requirements translate to:

  [[!plus#(O(_x0), O(_x1))]] = x1 >= x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(O(_x0), I(_x1))]] = 2x1 >= x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(I(_x0), O(_x1))]] = x1 >= x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(I(_x0), I(_x1))]] = 2x1 >= 0 = [[!plus#(!plus(_x0, _x1), I(0))]] 
  [[!plus#(I(_x0), I(_x1))]] = 2x1 >= x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(_x0, !plus(_x1, _x2))]] = 2 + x2 + 2x1 > x2 = [[!plus#(!plus(_x0, _x1), _x2)]] 
  [[!plus#(_x0, !plus(_x1, _x2))]] = 2 + x2 + 2x1 > x1 = [[!plus#(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_6, R_0, minimal, formative), where P_6 consists of:

  !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
  !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
  !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
  !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
  !plus#(I(X), I(Y)) =#> !plus#(X, Y)   

Thus, the original system is terminating if (P_6, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_6, R_0, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_6, R_0) are:

  O(0) => 0 
  !plus(0, X) => X 
  !plus(X, 0) => X 
  !plus(O(X), O(Y)) => O(!plus(X, Y)) 
  !plus(O(X), I(Y)) => I(!plus(X, Y)) 
  !plus(I(X), O(Y)) => I(!plus(X, Y)) 
  !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  !plus#(O(X), O(Y)) >? !plus#(X, Y) 
  !plus#(O(X), I(Y)) >? !plus#(X, Y) 
  !plus#(I(X), O(Y)) >? !plus#(X, Y) 
  !plus#(I(X), I(Y)) >? !plus#(!plus(X, Y), I(0)) 
  !plus#(I(X), I(Y)) >? !plus#(X, Y) 
  O(0) >= 0 
  !plus(0, X) >= X 
  !plus(X, 0) >= X 
  !plus(O(X), O(Y)) >= O(!plus(X, Y)) 
  !plus(O(X), I(Y)) >= I(!plus(X, Y)) 
  !plus(I(X), O(Y)) >= I(!plus(X, Y)) 
  !plus(I(X), I(Y)) >= O(!plus(!plus(X, Y), I(0))) 
  !plus(X, !plus(Y, Z)) >= !plus(!plus(X, Y), Z) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  !plus# = \y0y1.y0 + y1 
  0 = 0 
  I = \y0.1 + y0 
  O = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[!plus#(O(_x0), O(_x1))]] = 2 + x0 + x1 > x0 + x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(O(_x0), I(_x1))]] = 2 + x0 + x1 > x0 + x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(I(_x0), O(_x1))]] = 2 + x0 + x1 > x0 + x1 = [[!plus#(_x0, _x1)]] 
  [[!plus#(I(_x0), I(_x1))]] = 2 + x0 + x1 > 1 + x0 + x1 = [[!plus#(!plus(_x0, _x1), I(0))]] 
  [[!plus#(I(_x0), I(_x1))]] = 2 + x0 + x1 > x0 + x1 = [[!plus#(_x0, _x1)]] 
  [[O(0)]] = 1 >= 0 = [[0]] 
  [[!plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(O(_x0), O(_x1))]] = 2 + x0 + x1 >= 1 + x0 + x1 = [[O(!plus(_x0, _x1))]] 
  [[!plus(O(_x0), I(_x1))]] = 2 + x0 + x1 >= 1 + x0 + x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), O(_x1))]] = 2 + x0 + x1 >= 1 + x0 + x1 = [[I(!plus(_x0, _x1))]] 
  [[!plus(I(_x0), I(_x1))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[O(!plus(!plus(_x0, _x1), I(0)))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(!plus(_x0, _x1), _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, R_0) by ({}, R_0).  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.