/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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WORST_CASE(Omega(n^1), ?)
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) CpxTRS
(1) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(2) CpxTRS
(3) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
(4) typed CpxTrs
(5) OrderProof [LOWER BOUND(ID), 0 ms]
(6) typed CpxTrs
(7) RewriteLemmaProof [LOWER BOUND(ID), 322 ms]
(8) BEST
    (9) proven lower bound
        (10) LowerBoundPropagationProof [FINISHED, 0 ms]
        (11) BOUNDS(n^1, INF)
    (12) typed CpxTrs
        (13) RewriteLemmaProof [LOWER BOUND(ID), 91 ms]
        (14) typed CpxTrs
        (15) RewriteLemmaProof [LOWER BOUND(ID), 412 ms]
        (16) typed CpxTrs


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

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


The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(1) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
----------------------------------------

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


The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0', y) -> 0'
   minus(x, 0') -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0', y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0') -> true
   p(s(x)) -> x
   p(0') -> 0'
   div(x, y) -> quot(x, y, 0')
   quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
   quot(0', s(y), z) -> z
   ack(0', x) -> s(x)
   ack(0', x) -> plus(x, s(0'))
   ack(s(x), 0') -> ack(x, s(0'))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(3) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
----------------------------------------

(4)
Obligation:
TRS:
Rules:
minus(minus(x, y), z) -> minus(x, plus(y, z))
minus(0', y) -> 0'
minus(x, 0') -> x
minus(s(x), s(y)) -> minus(x, y)
plus(0', y) -> y
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(y, x))
zero(s(x)) -> false
zero(0') -> true
p(s(x)) -> x
p(0') -> 0'
div(x, y) -> quot(x, y, 0')
quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
quot(0', s(y), z) -> z
ack(0', x) -> s(x)
ack(0', x) -> plus(x, s(0'))
ack(s(x), 0') -> ack(x, s(0'))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Types:
minus :: 0':s -> 0':s -> 0':s
plus :: 0':s -> 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
zero :: 0':s -> false:true
false :: false:true
true :: false:true
p :: 0':s -> 0':s
div :: 0':s -> 0':s -> 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
ack :: 0':s -> 0':s -> 0':s
hole_0':s1_0 :: 0':s
hole_false:true2_0 :: false:true
gen_0':s3_0 :: Nat -> 0':s

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

(5) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
minus, plus, quot, ack

They will be analysed ascendingly in the following order:
plus < minus
minus < quot
plus < ack
ack < quot

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

(6)
Obligation:
TRS:
Rules:
minus(minus(x, y), z) -> minus(x, plus(y, z))
minus(0', y) -> 0'
minus(x, 0') -> x
minus(s(x), s(y)) -> minus(x, y)
plus(0', y) -> y
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(y, x))
zero(s(x)) -> false
zero(0') -> true
p(s(x)) -> x
p(0') -> 0'
div(x, y) -> quot(x, y, 0')
quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
quot(0', s(y), z) -> z
ack(0', x) -> s(x)
ack(0', x) -> plus(x, s(0'))
ack(s(x), 0') -> ack(x, s(0'))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Types:
minus :: 0':s -> 0':s -> 0':s
plus :: 0':s -> 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
zero :: 0':s -> false:true
false :: false:true
true :: false:true
p :: 0':s -> 0':s
div :: 0':s -> 0':s -> 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
ack :: 0':s -> 0':s -> 0':s
hole_0':s1_0 :: 0':s
hole_false:true2_0 :: false:true
gen_0':s3_0 :: Nat -> 0':s


Generator Equations:
gen_0':s3_0(0) <=> 0'
gen_0':s3_0(+(x, 1)) <=> s(gen_0':s3_0(x))


The following defined symbols remain to be analysed:
plus, minus, quot, ack

They will be analysed ascendingly in the following order:
plus < minus
minus < quot
plus < ack
ack < quot

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

(7) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
plus(gen_0':s3_0(n5_0), gen_0':s3_0(b)) -> gen_0':s3_0(+(n5_0, b)), rt in Omega(1 + n5_0)

Induction Base:
plus(gen_0':s3_0(0), gen_0':s3_0(b)) ->_R^Omega(1)
gen_0':s3_0(b)

Induction Step:
plus(gen_0':s3_0(+(n5_0, 1)), gen_0':s3_0(b)) ->_R^Omega(1)
plus(gen_0':s3_0(n5_0), s(gen_0':s3_0(b))) ->_IH
gen_0':s3_0(+(+(b, 1), c6_0))

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(8)
Complex Obligation (BEST)

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

(9)
Obligation:
Proved the lower bound n^1 for the following obligation:

TRS:
Rules:
minus(minus(x, y), z) -> minus(x, plus(y, z))
minus(0', y) -> 0'
minus(x, 0') -> x
minus(s(x), s(y)) -> minus(x, y)
plus(0', y) -> y
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(y, x))
zero(s(x)) -> false
zero(0') -> true
p(s(x)) -> x
p(0') -> 0'
div(x, y) -> quot(x, y, 0')
quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
quot(0', s(y), z) -> z
ack(0', x) -> s(x)
ack(0', x) -> plus(x, s(0'))
ack(s(x), 0') -> ack(x, s(0'))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Types:
minus :: 0':s -> 0':s -> 0':s
plus :: 0':s -> 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
zero :: 0':s -> false:true
false :: false:true
true :: false:true
p :: 0':s -> 0':s
div :: 0':s -> 0':s -> 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
ack :: 0':s -> 0':s -> 0':s
hole_0':s1_0 :: 0':s
hole_false:true2_0 :: false:true
gen_0':s3_0 :: Nat -> 0':s


Generator Equations:
gen_0':s3_0(0) <=> 0'
gen_0':s3_0(+(x, 1)) <=> s(gen_0':s3_0(x))


The following defined symbols remain to be analysed:
plus, minus, quot, ack

They will be analysed ascendingly in the following order:
plus < minus
minus < quot
plus < ack
ack < quot

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

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

(11)
BOUNDS(n^1, INF)

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

(12)
Obligation:
TRS:
Rules:
minus(minus(x, y), z) -> minus(x, plus(y, z))
minus(0', y) -> 0'
minus(x, 0') -> x
minus(s(x), s(y)) -> minus(x, y)
plus(0', y) -> y
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(y, x))
zero(s(x)) -> false
zero(0') -> true
p(s(x)) -> x
p(0') -> 0'
div(x, y) -> quot(x, y, 0')
quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
quot(0', s(y), z) -> z
ack(0', x) -> s(x)
ack(0', x) -> plus(x, s(0'))
ack(s(x), 0') -> ack(x, s(0'))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Types:
minus :: 0':s -> 0':s -> 0':s
plus :: 0':s -> 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
zero :: 0':s -> false:true
false :: false:true
true :: false:true
p :: 0':s -> 0':s
div :: 0':s -> 0':s -> 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
ack :: 0':s -> 0':s -> 0':s
hole_0':s1_0 :: 0':s
hole_false:true2_0 :: false:true
gen_0':s3_0 :: Nat -> 0':s


Lemmas:
plus(gen_0':s3_0(n5_0), gen_0':s3_0(b)) -> gen_0':s3_0(+(n5_0, b)), rt in Omega(1 + n5_0)


Generator Equations:
gen_0':s3_0(0) <=> 0'
gen_0':s3_0(+(x, 1)) <=> s(gen_0':s3_0(x))


The following defined symbols remain to be analysed:
minus, quot, ack

They will be analysed ascendingly in the following order:
minus < quot
ack < quot

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

(13) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
minus(gen_0':s3_0(n870_0), gen_0':s3_0(n870_0)) -> gen_0':s3_0(0), rt in Omega(1 + n870_0)

Induction Base:
minus(gen_0':s3_0(0), gen_0':s3_0(0)) ->_R^Omega(1)
0'

Induction Step:
minus(gen_0':s3_0(+(n870_0, 1)), gen_0':s3_0(+(n870_0, 1))) ->_R^Omega(1)
minus(gen_0':s3_0(n870_0), gen_0':s3_0(n870_0)) ->_IH
gen_0':s3_0(0)

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(14)
Obligation:
TRS:
Rules:
minus(minus(x, y), z) -> minus(x, plus(y, z))
minus(0', y) -> 0'
minus(x, 0') -> x
minus(s(x), s(y)) -> minus(x, y)
plus(0', y) -> y
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(y, x))
zero(s(x)) -> false
zero(0') -> true
p(s(x)) -> x
p(0') -> 0'
div(x, y) -> quot(x, y, 0')
quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
quot(0', s(y), z) -> z
ack(0', x) -> s(x)
ack(0', x) -> plus(x, s(0'))
ack(s(x), 0') -> ack(x, s(0'))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Types:
minus :: 0':s -> 0':s -> 0':s
plus :: 0':s -> 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
zero :: 0':s -> false:true
false :: false:true
true :: false:true
p :: 0':s -> 0':s
div :: 0':s -> 0':s -> 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
ack :: 0':s -> 0':s -> 0':s
hole_0':s1_0 :: 0':s
hole_false:true2_0 :: false:true
gen_0':s3_0 :: Nat -> 0':s


Lemmas:
plus(gen_0':s3_0(n5_0), gen_0':s3_0(b)) -> gen_0':s3_0(+(n5_0, b)), rt in Omega(1 + n5_0)
minus(gen_0':s3_0(n870_0), gen_0':s3_0(n870_0)) -> gen_0':s3_0(0), rt in Omega(1 + n870_0)


Generator Equations:
gen_0':s3_0(0) <=> 0'
gen_0':s3_0(+(x, 1)) <=> s(gen_0':s3_0(x))


The following defined symbols remain to be analysed:
ack, quot

They will be analysed ascendingly in the following order:
ack < quot

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

(15) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
ack(gen_0':s3_0(1), gen_0':s3_0(+(1, n1304_0))) -> *4_0, rt in Omega(n1304_0)

Induction Base:
ack(gen_0':s3_0(1), gen_0':s3_0(+(1, 0)))

Induction Step:
ack(gen_0':s3_0(1), gen_0':s3_0(+(1, +(n1304_0, 1)))) ->_R^Omega(1)
ack(gen_0':s3_0(0), ack(s(gen_0':s3_0(0)), gen_0':s3_0(+(1, n1304_0)))) ->_IH
ack(gen_0':s3_0(0), *4_0)

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(16)
Obligation:
TRS:
Rules:
minus(minus(x, y), z) -> minus(x, plus(y, z))
minus(0', y) -> 0'
minus(x, 0') -> x
minus(s(x), s(y)) -> minus(x, y)
plus(0', y) -> y
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(y, x))
zero(s(x)) -> false
zero(0') -> true
p(s(x)) -> x
p(0') -> 0'
div(x, y) -> quot(x, y, 0')
quot(s(x), s(y), z) -> quot(minus(p(ack(0', x)), y), s(y), s(z))
quot(0', s(y), z) -> z
ack(0', x) -> s(x)
ack(0', x) -> plus(x, s(0'))
ack(s(x), 0') -> ack(x, s(0'))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Types:
minus :: 0':s -> 0':s -> 0':s
plus :: 0':s -> 0':s -> 0':s
0' :: 0':s
s :: 0':s -> 0':s
zero :: 0':s -> false:true
false :: false:true
true :: false:true
p :: 0':s -> 0':s
div :: 0':s -> 0':s -> 0':s
quot :: 0':s -> 0':s -> 0':s -> 0':s
ack :: 0':s -> 0':s -> 0':s
hole_0':s1_0 :: 0':s
hole_false:true2_0 :: false:true
gen_0':s3_0 :: Nat -> 0':s


Lemmas:
plus(gen_0':s3_0(n5_0), gen_0':s3_0(b)) -> gen_0':s3_0(+(n5_0, b)), rt in Omega(1 + n5_0)
minus(gen_0':s3_0(n870_0), gen_0':s3_0(n870_0)) -> gen_0':s3_0(0), rt in Omega(1 + n870_0)
ack(gen_0':s3_0(1), gen_0':s3_0(+(1, n1304_0))) -> *4_0, rt in Omega(n1304_0)


Generator Equations:
gen_0':s3_0(0) <=> 0'
gen_0':s3_0(+(x, 1)) <=> s(gen_0':s3_0(x))


The following defined symbols remain to be analysed:
quot