TRS Standard pair #516966949

details

property	value
status	complete
benchmark	4.32.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n084.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	2.22941684723 seconds
cpu usage	7.458075266
max memory	6.5073152E8

stage attributes

key	value
output-size	2719
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES

Problem:
 a(b(x)) -> b(a(a(x)))
 b(c(x)) -> c(b(b(x)))
 c(a(x)) -> a(c(c(x)))
 u(a(x)) -> x
 v(b(x)) -> x
 w(c(x)) -> x
 a(u(x)) -> x
 b(v(x)) -> x
 c(w(x)) -> x

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [v](x0) = 4x0,
   
   [a](x0) = x0,
   
   [u](x0) = 2x0 + 4,
   
   [w](x0) = 8x0,
   
   [b](x0) = x0,
   
   [c](x0) = x0
  orientation:
   a(b(x)) = x >= x = b(a(a(x)))
   
   b(c(x)) = x >= x = c(b(b(x)))
   
   c(a(x)) = x >= x = a(c(c(x)))
   
   u(a(x)) = 2x + 4 >= x = x
   
   v(b(x)) = 4x >= x = x
   
   w(c(x)) = 8x >= x = x
   
   a(u(x)) = 2x + 4 >= x = x
   
   b(v(x)) = 4x >= x = x
   
   c(w(x)) = 8x >= x = x
  problem:
   a(b(x)) -> b(a(a(x)))
   b(c(x)) -> c(b(b(x)))
   c(a(x)) -> a(c(c(x)))
   v(b(x)) -> x
   w(c(x)) -> x
   b(v(x)) -> x
   c(w(x)) -> x
  String Reversal Processor:
   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))
   b(v(x)) -> x
   c(w(x)) -> x
   v(b(x)) -> x
   w(c(x)) -> x
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [v](x0) = x0 + 2,
     
     [a](x0) = x0,
     
     [w](x0) = 2x0,
     
     [b](x0) = x0,
     
     [c](x0) = x0
    orientation:
     b(a(x)) = x >= x = a(a(b(x)))
     
     c(b(x)) = x >= x = b(b(c(x)))
     
     a(c(x)) = x >= x = c(c(a(x)))
     
     b(v(x)) = x + 2 >= x = x
     
     c(w(x)) = 2x >= x = x
     
     v(b(x)) = x + 2 >= x = x
     
     w(c(x)) = 2x >= x = x
    problem:
     b(a(x)) -> a(a(b(x)))
     c(b(x)) -> b(b(c(x)))
     a(c(x)) -> c(c(a(x)))
     c(w(x)) -> x
     w(c(x)) -> x

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