WORST_CASE(Omega(n^1),?)
proof of /export/starexec/sandbox/benchmark/theBenchmark.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).

(0) CpxRelTRS
(1) SInnermostTerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 1697 ms]
(2) CpxRelTRS
(3) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(4) TRS for Loop Detection
(5) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
(6) BEST
    (7) proven lower bound
        (8) LowerBoundPropagationProof [FINISHED, 0 ms]
        (9) BOUNDS(n^1, INF)
    (10) TRS for Loop Detection


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

(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   LAST(nil) -> c
   LAST(cons(z0, nil)) -> c1
   LAST(cons(z0, cons(z1, z2))) -> c2(LAST(cons(z1, z2)))
   DEL(z0, nil) -> c3
   DEL(z0, cons(z1, z2)) -> c4(IF(eq(z0, z1), z0, z1, z2), EQ(z0, z1))
   IF(true, z0, z1, z2) -> c5
   IF(false, z0, z1, z2) -> c6(DEL(z0, z2))
   EQ(0, 0) -> c7
   EQ(0, s(z0)) -> c8
   EQ(s(z0), 0) -> c9
   EQ(s(z0), s(z1)) -> c10(EQ(z0, z1))
   REVERSE(nil) -> c11
   REVERSE(cons(z0, z1)) -> c12(LAST(cons(z0, z1)))
   REVERSE(cons(z0, z1)) -> c13(REVERSE(del(last(cons(z0, z1)), cons(z0, z1))), DEL(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1)))

The (relative) TRS S consists of the following rules:

   last(nil) -> 0
   last(cons(z0, nil)) -> z0
   last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2))
   del(z0, nil) -> nil
   del(z0, cons(z1, z2)) -> if(eq(z0, z1), z0, z1, z2)
   if(true, z0, z1, z2) -> z2
   if(false, z0, z1, z2) -> cons(z1, del(z0, z2))
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   reverse(nil) -> nil
   reverse(cons(z0, z1)) -> cons(last(cons(z0, z1)), reverse(del(last(cons(z0, z1)), cons(z0, z1))))

Rewrite Strategy: INNERMOST
----------------------------------------

(1) SInnermostTerminationProof (BOTH CONCRETE BOUNDS(ID, ID))
proved innermost termination of relative rules
----------------------------------------

(2)
Obligation:
The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   LAST(nil) -> c
   LAST(cons(z0, nil)) -> c1
   LAST(cons(z0, cons(z1, z2))) -> c2(LAST(cons(z1, z2)))
   DEL(z0, nil) -> c3
   DEL(z0, cons(z1, z2)) -> c4(IF(eq(z0, z1), z0, z1, z2), EQ(z0, z1))
   IF(true, z0, z1, z2) -> c5
   IF(false, z0, z1, z2) -> c6(DEL(z0, z2))
   EQ(0, 0) -> c7
   EQ(0, s(z0)) -> c8
   EQ(s(z0), 0) -> c9
   EQ(s(z0), s(z1)) -> c10(EQ(z0, z1))
   REVERSE(nil) -> c11
   REVERSE(cons(z0, z1)) -> c12(LAST(cons(z0, z1)))
   REVERSE(cons(z0, z1)) -> c13(REVERSE(del(last(cons(z0, z1)), cons(z0, z1))), DEL(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1)))

The (relative) TRS S consists of the following rules:

   last(nil) -> 0
   last(cons(z0, nil)) -> z0
   last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2))
   del(z0, nil) -> nil
   del(z0, cons(z1, z2)) -> if(eq(z0, z1), z0, z1, z2)
   if(true, z0, z1, z2) -> z2
   if(false, z0, z1, z2) -> cons(z1, del(z0, z2))
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   reverse(nil) -> nil
   reverse(cons(z0, z1)) -> cons(last(cons(z0, z1)), reverse(del(last(cons(z0, z1)), cons(z0, z1))))

Rewrite Strategy: INNERMOST
----------------------------------------

(3) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
----------------------------------------

(4)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   LAST(nil) -> c
   LAST(cons(z0, nil)) -> c1
   LAST(cons(z0, cons(z1, z2))) -> c2(LAST(cons(z1, z2)))
   DEL(z0, nil) -> c3
   DEL(z0, cons(z1, z2)) -> c4(IF(eq(z0, z1), z0, z1, z2), EQ(z0, z1))
   IF(true, z0, z1, z2) -> c5
   IF(false, z0, z1, z2) -> c6(DEL(z0, z2))
   EQ(0, 0) -> c7
   EQ(0, s(z0)) -> c8
   EQ(s(z0), 0) -> c9
   EQ(s(z0), s(z1)) -> c10(EQ(z0, z1))
   REVERSE(nil) -> c11
   REVERSE(cons(z0, z1)) -> c12(LAST(cons(z0, z1)))
   REVERSE(cons(z0, z1)) -> c13(REVERSE(del(last(cons(z0, z1)), cons(z0, z1))), DEL(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1)))

The (relative) TRS S consists of the following rules:

   last(nil) -> 0
   last(cons(z0, nil)) -> z0
   last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2))
   del(z0, nil) -> nil
   del(z0, cons(z1, z2)) -> if(eq(z0, z1), z0, z1, z2)
   if(true, z0, z1, z2) -> z2
   if(false, z0, z1, z2) -> cons(z1, del(z0, z2))
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   reverse(nil) -> nil
   reverse(cons(z0, z1)) -> cons(last(cons(z0, z1)), reverse(del(last(cons(z0, z1)), cons(z0, z1))))

Rewrite Strategy: INNERMOST
----------------------------------------

(5) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

LAST(cons(z0, cons(z1, z2))) ->^+ c2(LAST(cons(z1, z2)))

gives rise to a decreasing loop by considering the right hand sides subterm at position [0].

The pumping substitution is [z2 / cons(z1, z2)].

The result substitution is [z0 / z1].




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

(6)
Complex Obligation (BEST)

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

(7)
Obligation:
Proved the lower bound n^1 for the following obligation:

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   LAST(nil) -> c
   LAST(cons(z0, nil)) -> c1
   LAST(cons(z0, cons(z1, z2))) -> c2(LAST(cons(z1, z2)))
   DEL(z0, nil) -> c3
   DEL(z0, cons(z1, z2)) -> c4(IF(eq(z0, z1), z0, z1, z2), EQ(z0, z1))
   IF(true, z0, z1, z2) -> c5
   IF(false, z0, z1, z2) -> c6(DEL(z0, z2))
   EQ(0, 0) -> c7
   EQ(0, s(z0)) -> c8
   EQ(s(z0), 0) -> c9
   EQ(s(z0), s(z1)) -> c10(EQ(z0, z1))
   REVERSE(nil) -> c11
   REVERSE(cons(z0, z1)) -> c12(LAST(cons(z0, z1)))
   REVERSE(cons(z0, z1)) -> c13(REVERSE(del(last(cons(z0, z1)), cons(z0, z1))), DEL(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1)))

The (relative) TRS S consists of the following rules:

   last(nil) -> 0
   last(cons(z0, nil)) -> z0
   last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2))
   del(z0, nil) -> nil
   del(z0, cons(z1, z2)) -> if(eq(z0, z1), z0, z1, z2)
   if(true, z0, z1, z2) -> z2
   if(false, z0, z1, z2) -> cons(z1, del(z0, z2))
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   reverse(nil) -> nil
   reverse(cons(z0, z1)) -> cons(last(cons(z0, z1)), reverse(del(last(cons(z0, z1)), cons(z0, z1))))

Rewrite Strategy: INNERMOST
----------------------------------------

(8) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(9)
BOUNDS(n^1, INF)

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

(10)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   LAST(nil) -> c
   LAST(cons(z0, nil)) -> c1
   LAST(cons(z0, cons(z1, z2))) -> c2(LAST(cons(z1, z2)))
   DEL(z0, nil) -> c3
   DEL(z0, cons(z1, z2)) -> c4(IF(eq(z0, z1), z0, z1, z2), EQ(z0, z1))
   IF(true, z0, z1, z2) -> c5
   IF(false, z0, z1, z2) -> c6(DEL(z0, z2))
   EQ(0, 0) -> c7
   EQ(0, s(z0)) -> c8
   EQ(s(z0), 0) -> c9
   EQ(s(z0), s(z1)) -> c10(EQ(z0, z1))
   REVERSE(nil) -> c11
   REVERSE(cons(z0, z1)) -> c12(LAST(cons(z0, z1)))
   REVERSE(cons(z0, z1)) -> c13(REVERSE(del(last(cons(z0, z1)), cons(z0, z1))), DEL(last(cons(z0, z1)), cons(z0, z1)), LAST(cons(z0, z1)))

The (relative) TRS S consists of the following rules:

   last(nil) -> 0
   last(cons(z0, nil)) -> z0
   last(cons(z0, cons(z1, z2))) -> last(cons(z1, z2))
   del(z0, nil) -> nil
   del(z0, cons(z1, z2)) -> if(eq(z0, z1), z0, z1, z2)
   if(true, z0, z1, z2) -> z2
   if(false, z0, z1, z2) -> cons(z1, del(z0, z2))
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   reverse(nil) -> nil
   reverse(cons(z0, z1)) -> cons(last(cons(z0, z1)), reverse(del(last(cons(z0, z1)), cons(z0, z1))))

Rewrite Strategy: INNERMOST