TRS Standard pair #615297811

details
property	value
status	complete
benchmark	sat.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n050.star.cs.uiowa.edu
space	TCT_12
run statistics
property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.32264995575 seconds
cpu usage	0.432125504
max memory	3.8793216E7
stage attributes
key	value
output-size	5562
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...
## 5 initial DP problems to solve.
## First, we try to decompose these problems into smaller problems.
## Round 1 [5 DP problems]:
  ## DP problem:
  Dependency pairs = [verify^#(cons(_0,_1)) -> verify^#(_1)]
  TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Success, finiteness proved!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [guess^#(cons(_0,_1)) -> guess^#(_1)]
  TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Success, finiteness proved!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [choice^#(cons(_0,_1)) -> choice^#(_1)]
  TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Success, finiteness proved!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [member^#(_0,cons(_1,_2)) -> member^#(_0,_2)]
  TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Success, finiteness proved!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [eq^#(O(_0),0(_1)) -> eq^#(_0,_1), eq^#(1(_0),1(_1)) -> eq^#(_0,_1)]
  TRS = {if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, member(_0,nil) -> false, member(_0,cons(_1,_2)) -> if(eq(_0,_1),true,member(_0,_2)), eq(nil,nil) -> true, eq(O(_0),0(_1)) -> eq(_0,_1), eq(0(_0),1(_1)) -> false, eq(1(_0),0(_1)) -> false, eq(1(_0),1(_1)) -> eq(_0,_1), negate(0(_0)) -> 1(_0), negate(1(_0)) -> 0(_0), choice(cons(_0,_1)) -> _0, choice(cons(_0,_1)) -> choice(_1), guess(nil) -> nil, guess(cons(_0,_1)) -> cons(choice(_0),guess(_1)), verify(nil) -> true, verify(cons(_0,_1)) -> if(member(negate(_0),_1),false,verify(_1)), sat(_0) -> satck(_0,guess(_0)), satck(_0,_1) -> if(verify(_1),_1,unsat)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      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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