Problem Transformed CSR 04 Ex15 Luc98 C

Tool CaT

Execution Time	Unknown
Answer	MAYBE
Input	Transformed CSR 04 Ex15 Luc98 C

stdout:

MAYBE

Problem:
active(and(true(),X)) -> mark(X)
active(and(false(),Y)) -> mark(false())
active(if(true(),X,Y)) -> mark(X)
active(if(false(),X,Y)) -> mark(Y)
active(add(0(),X)) -> mark(X)
active(add(s(X),Y)) -> mark(s(add(X,Y)))
active(first(0(),X)) -> mark(nil())
active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
active(from(X)) -> mark(cons(X,from(s(X))))
active(and(X1,X2)) -> and(active(X1),X2)
active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
active(add(X1,X2)) -> add(active(X1),X2)
active(first(X1,X2)) -> first(active(X1),X2)
active(first(X1,X2)) -> first(X1,active(X2))
and(mark(X1),X2) -> mark(and(X1,X2))
if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
add(mark(X1),X2) -> mark(add(X1,X2))
first(mark(X1),X2) -> mark(first(X1,X2))
first(X1,mark(X2)) -> mark(first(X1,X2))
proper(and(X1,X2)) -> and(proper(X1),proper(X2))
proper(true()) -> ok(true())
proper(false()) -> ok(false())
proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
proper(add(X1,X2)) -> add(proper(X1),proper(X2))
proper(0()) -> ok(0())
proper(s(X)) -> s(proper(X))
proper(first(X1,X2)) -> first(proper(X1),proper(X2))
proper(nil()) -> ok(nil())
proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
proper(from(X)) -> from(proper(X))
and(ok(X1),ok(X2)) -> ok(and(X1,X2))
if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
add(ok(X1),ok(X2)) -> ok(add(X1,X2))
s(ok(X)) -> ok(s(X))
first(ok(X1),ok(X2)) -> ok(first(X1,X2))
cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
from(ok(X)) -> ok(from(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Proof:
Open

Tool IRC1

Execution Time	Unknown
Answer	MAYBE
Input	Transformed CSR 04 Ex15 Luc98 C

stdout:

MAYBE

Tool IRC2

Execution Time	Unknown
Answer	TIMEOUT
Input	Transformed CSR 04 Ex15 Luc98 C

stdout:

TIMEOUT

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           TIMEOUT
Input Problem:    innermost runtime-complexity with respect to
  Rules:
    {  active(and(true(), X)) -> mark(X)
     , active(and(false(), Y)) -> mark(false())
     , active(if(true(), X, Y)) -> mark(X)
     , active(if(false(), X, Y)) -> mark(Y)
     , active(add(0(), X)) -> mark(X)
     , active(add(s(X), Y)) -> mark(s(add(X, Y)))
     , active(first(0(), X)) -> mark(nil())
     , active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
     , active(from(X)) -> mark(cons(X, from(s(X))))
     , active(and(X1, X2)) -> and(active(X1), X2)
     , active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
     , active(add(X1, X2)) -> add(active(X1), X2)
     , active(first(X1, X2)) -> first(active(X1), X2)
     , active(first(X1, X2)) -> first(X1, active(X2))
     , and(mark(X1), X2) -> mark(and(X1, X2))
     , if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
     , add(mark(X1), X2) -> mark(add(X1, X2))
     , first(mark(X1), X2) -> mark(first(X1, X2))
     , first(X1, mark(X2)) -> mark(first(X1, X2))
     , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
     , proper(true()) -> ok(true())
     , proper(false()) -> ok(false())
     , proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
     , proper(add(X1, X2)) -> add(proper(X1), proper(X2))
     , proper(0()) -> ok(0())
     , proper(s(X)) -> s(proper(X))
     , proper(first(X1, X2)) -> first(proper(X1), proper(X2))
     , proper(nil()) -> ok(nil())
     , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
     , proper(from(X)) -> from(proper(X))
     , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
     , if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
     , add(ok(X1), ok(X2)) -> ok(add(X1, X2))
     , s(ok(X)) -> ok(s(X))
     , first(ok(X1), ok(X2)) -> ok(first(X1, X2))
     , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
     , from(ok(X)) -> ok(from(X))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}

Proof Output:
  Computation stopped due to timeout after 60.0 seconds

Tool RC1

Execution Time	Unknown
Answer	MAYBE
Input	Transformed CSR 04 Ex15 Luc98 C

stdout:

MAYBE

Tool RC2

Execution Time	Unknown
Answer	TIMEOUT
Input	Transformed CSR 04 Ex15 Luc98 C

stdout:

TIMEOUT

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           TIMEOUT
Input Problem:    runtime-complexity with respect to
  Rules:
    {  active(and(true(), X)) -> mark(X)
     , active(and(false(), Y)) -> mark(false())
     , active(if(true(), X, Y)) -> mark(X)
     , active(if(false(), X, Y)) -> mark(Y)
     , active(add(0(), X)) -> mark(X)
     , active(add(s(X), Y)) -> mark(s(add(X, Y)))
     , active(first(0(), X)) -> mark(nil())
     , active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
     , active(from(X)) -> mark(cons(X, from(s(X))))
     , active(and(X1, X2)) -> and(active(X1), X2)
     , active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
     , active(add(X1, X2)) -> add(active(X1), X2)
     , active(first(X1, X2)) -> first(active(X1), X2)
     , active(first(X1, X2)) -> first(X1, active(X2))
     , and(mark(X1), X2) -> mark(and(X1, X2))
     , if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
     , add(mark(X1), X2) -> mark(add(X1, X2))
     , first(mark(X1), X2) -> mark(first(X1, X2))
     , first(X1, mark(X2)) -> mark(first(X1, X2))
     , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
     , proper(true()) -> ok(true())
     , proper(false()) -> ok(false())
     , proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
     , proper(add(X1, X2)) -> add(proper(X1), proper(X2))
     , proper(0()) -> ok(0())
     , proper(s(X)) -> s(proper(X))
     , proper(first(X1, X2)) -> first(proper(X1), proper(X2))
     , proper(nil()) -> ok(nil())
     , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
     , proper(from(X)) -> from(proper(X))
     , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
     , if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
     , add(ok(X1), ok(X2)) -> ok(add(X1, X2))
     , s(ok(X)) -> ok(s(X))
     , first(ok(X1), ok(X2)) -> ok(first(X1, X2))
     , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
     , from(ok(X)) -> ok(from(X))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}

Proof Output:
  Computation stopped due to timeout after 60.0 seconds