TRS Standard pair #516964174

details

property	value
status	complete
benchmark	17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n034.star.cs.uiowa.edu
space	Der95

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	0.523255109787 seconds
cpu usage	0.901544333
max memory	1.68976384E8

stage attributes

key	value
output-size	1855
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:
 .(1(),x) -> x
 .(x,1()) -> x
 .(i(x),x) -> 1()
 .(x,i(x)) -> 1()
 i(1()) -> 1()
 i(i(x)) -> x
 .(i(y),.(y,z)) -> z
 .(y,.(i(y),z)) -> z
 .(.(x,y),z) -> .(x,.(y,z))
 i(.(x,y)) -> .(i(y),i(x))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [.](x0, x1) = x0 + x1 + 2,
   
   [1] = 1,
   
   [i](x0) = x0
  orientation:
   .(1(),x) = x + 3 >= x = x
   
   .(x,1()) = x + 3 >= x = x
   
   .(i(x),x) = 2x + 2 >= 1 = 1()
   
   .(x,i(x)) = 2x + 2 >= 1 = 1()
   
   i(1()) = 1 >= 1 = 1()
   
   i(i(x)) = x >= x = x
   
   .(i(y),.(y,z)) = 2y + z + 4 >= z = z
   
   .(y,.(i(y),z)) = 2y + z + 4 >= z = z
   
   .(.(x,y),z) = x + y + z + 4 >= x + y + z + 4 = .(x,.(y,z))
   
   i(.(x,y)) = x + y + 2 >= x + y + 2 = .(i(y),i(x))
  problem:
   i(1()) -> 1()
   i(i(x)) -> x
   .(.(x,y),z) -> .(x,.(y,z))
   i(.(x,y)) -> .(i(y),i(x))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [.](x0, x1) = x0 + x1 + 2,
    
    [1] = 0,
    
    [i](x0) = 4x0
   orientation:
    i(1()) = 0 >= 0 = 1()
    
    i(i(x)) = 16x >= x = x
    
    .(.(x,y),z) = x + y + z + 4 >= x + y + z + 4 = .(x,.(y,z))
    
    i(.(x,y)) = 4x + 4y + 8 >= 4x + 4y + 2 = .(i(y),i(x))
   problem:
    i(1()) -> 1()
    i(i(x)) -> x
    .(.(x,y),z) -> .(x,.(y,z))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [.](x0, x1) = 4x0 + x1 + 4,
     
     [1] = 0,
     
     [i](x0) = 3x0 + 6
    orientation:
     i(1()) = 6 >= 0 = 1()
     
     i(i(x)) = 9x + 24 >= x = x
     
     .(.(x,y),z) = 16x + 4y + z + 20 >= 4x + 4y + z + 8 = .(x,.(y,z))
    problem:
     
    Qed

popout

output may be truncated. 'popout' for the full output.

job log

popout

actions all output return to TRS Standard