TRS Standard pair #615303047

details
property	value
status	complete
benchmark	ReverseLastInit.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n014.star.cs.uiowa.edu
space	Applicative_05
run statistics
property	value
solver	NTI_23
configuration	default
runtime (wallclock)	77.3710989952 seconds
cpu usage	163.860191051
max memory	5.38509312E8
stage attributes
key	value
output-size	5176
starexec-result	"YES"
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES

** BEGIN proof argument **
All the DP problems were proved finite.
As all the involved DP processors are sound,
the TRS under analysis terminates.
** END proof argument **

** BEGIN proof description **
## Searching for a generalized rewrite rule
(a rule whose right-hand side contains a variable
that does not occur in the left-hand side)...
No generalized rewrite rule found!

## Applying the DP framework...
## 1 initial DP problem to solve.
## First, we try to decompose this problem into smaller problems.
## Round 1 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [app^#(app(app(compose,_0),_1),_2) -> app^#(_1,app(_0,_2)), app^#(app(app(compose,_0),_1),_2) -> app^#(_0,_2), app^#(reverse,_0) -> app^#(app(reverse2,_0),nil), app^#(app(reverse2,app(app(cons,_0),_1)),_2) -> app^#(app(reverse2,_1),app(app(cons,_0),_2)), app^#(app(reverse2,app(app(cons,_0),_1)),_2) -> app^#(app(cons,_0),_2)]
  TRS = {app(app(app(compose,_0),_1),_2) -> app(_1,app(_0,_2)), app(reverse,_0) -> app(app(reverse2,_0),nil), app(app(reverse2,nil),_0) -> _0, app(app(reverse2,app(app(cons,_0),_1)),_2) -> app(app(reverse2,_1),app(app(cons,_0),_2)), app(hd,app(app(cons,_0),_1)) -> _0, app(tl,app(app(cons,_0),_1)) -> _1, last -> app(app(compose,hd),reverse), init -> app(app(compose,reverse),app(app(compose,tl),reverse))}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: lexicographic path orders. 
      Failed!
    ## DP Processor: polynomial interpretations. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Polynomial order induced by the interpretation: {nil:[1], init:[2], reverse:[1], cons:[2], reverse2:[1], last:[2], compose:[1], app^#(_0,_1):[2 + 2 * _0 * _1], hd:[2], tl:[2], app(_0,_1):[_0 * _1]}
      and the value mu = 1 (see [Giesl, RTA'95])
      Decomposed the DP problem into 1 smaller problem to solve!
## Round 2 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [app^#(reverse,_0) -> app^#(app(reverse2,_0),nil), app^#(app(reverse2,app(app(cons,_0),_1)),_2) -> app^#(app(reverse2,_1),app(app(cons,_0),_2)), app^#(app(reverse2,app(app(cons,_0),_1)),_2) -> app^#(app(cons,_0),_2)]
  TRS = {app(app(app(compose,_0),_1),_2) -> app(_1,app(_0,_2)), app(reverse,_0) -> app(app(reverse2,_0),nil), app(app(reverse2,nil),_0) -> _0, app(app(reverse2,app(app(cons,_0),_1)),_2) -> app(app(reverse2,_1),app(app(cons,_0),_2)), app(hd,app(app(cons,_0),_1)) -> _0, app(tl,app(app(cons,_0),_1)) -> _1, last -> app(app(compose,hd),reverse), init -> app(app(compose,reverse),app(app(compose,tl),reverse))}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: lexicographic path orders. 
      Failed!
    ## DP Processor: polynomial interpretations. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Polynomial order induced by the interpretation: {nil:[1], init:[2], reverse:[1], cons:[2], reverse2:[1], last:[2], compose:[1], app^#(_0,_1):[2 + 2 * _0 * _1], hd:[2], tl:[2], app(_0,_1):[_0 * _1]}
      and the value mu = 1 (see [Giesl, RTA'95])
      Decomposed the DP problem into 1 smaller problem to solve!
## Round 3 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [app^#(app(reverse2,app(app(cons,_0),_1)),_2) -> app^#(app(reverse2,_1),app(app(cons,_0),_2)), app^#(app(reverse2,app(app(cons,_0),_1)),_2) -> app^#(app(cons,_0),_2)]
  TRS = {app(app(app(compose,_0),_1),_2) -> app(_1,app(_0,_2)), app(reverse,_0) -> app(app(reverse2,_0),nil), app(app(reverse2,nil),_0) -> _0, app(app(reverse2,app(app(cons,_0),_1)),_2) -> app(app(reverse2,_1),app(app(cons,_0),_2)), app(hd,app(app(cons,_0),_1)) -> _0, app(tl,app(app(cons,_0),_1)) -> _1, last -> app(app(compose,hd),reverse), init -> app(app(compose,reverse),app(app(compose,tl),reverse))}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: lexicographic path orders. 
      Failed!
    ## DP Processor: polynomial interpretations. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Polynomial order induced by the interpretation: {nil:[0], init:[2], reverse:[1], cons:[1], reverse2:[1], last:[1], compose:[0], app^#(_0,_1):[_0], hd:[0], tl:[0], app(_0,_1):[_0 + _1]}
      and the value mu = 0 (see [Giesl, RTA'95])
      Success, finiteness proved!
  This DP problem is finite.
## All the DP problems were proved finite.
As all the involved DP processors are sound,
the TRS under analysis terminates.
Proof run on Linux version 3.10.0-1160.62.1.el7.x86_64 for amd64
using Java version 1.8.0_322
** END proof description **
Total number of generated unfolded rules = 0
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