SRS Standard pair #487091762

details

property	value
status	complete
benchmark	z075.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	3.55954 seconds
cpu usage	12.7975
user time	10.0317
system time	2.76578
max virtual memory	5941208.0
max residence set size	65584.0

stage attributes

key	value
starexec-result	YES

output

YES

Problem:
 t(f(x1)) -> t(c(n(x1)))
 n(f(x1)) -> f(n(x1))
 o(f(x1)) -> f(o(x1))
 n(s(x1)) -> f(s(x1))
 o(s(x1)) -> f(s(x1))
 c(f(x1)) -> f(c(x1))
 c(n(x1)) -> n(c(x1))
 c(o(x1)) -> o(c(x1))
 c(o(x1)) -> o(x1)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [o](x0) = x0 + 8,
   
   [t](x0) = x0,
   
   [c](x0) = x0,
   
   [s](x0) = 8x0,
   
   [f](x0) = x0,
   
   [n](x0) = x0
  orientation:
   t(f(x1)) = x1 >= x1 = t(c(n(x1)))
   
   n(f(x1)) = x1 >= x1 = f(n(x1))
   
   o(f(x1)) = x1 + 8 >= x1 + 8 = f(o(x1))
   
   n(s(x1)) = 8x1 >= 8x1 = f(s(x1))
   
   o(s(x1)) = 8x1 + 8 >= 8x1 = f(s(x1))
   
   c(f(x1)) = x1 >= x1 = f(c(x1))
   
   c(n(x1)) = x1 >= x1 = n(c(x1))
   
   c(o(x1)) = x1 + 8 >= x1 + 8 = o(c(x1))
   
   c(o(x1)) = x1 + 8 >= x1 + 8 = o(x1)
  problem:
   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [o](x0) = 6x0 + 1,
    
    [t](x0) = 2x0 + 1,
    
    [c](x0) = x0,
    
    [s](x0) = x0 + 1,
    
    [f](x0) = 4x0 + 1,
    
    [n](x0) = 4x0 + 1
   orientation:
    t(f(x1)) = 8x1 + 3 >= 8x1 + 3 = t(c(n(x1)))
    
    n(f(x1)) = 16x1 + 5 >= 16x1 + 5 = f(n(x1))
    
    o(f(x1)) = 24x1 + 7 >= 24x1 + 5 = f(o(x1))
    
    n(s(x1)) = 4x1 + 5 >= 4x1 + 5 = f(s(x1))
    
    c(f(x1)) = 4x1 + 1 >= 4x1 + 1 = f(c(x1))
    
    c(n(x1)) = 4x1 + 1 >= 4x1 + 1 = n(c(x1))
    
    c(o(x1)) = 6x1 + 1 >= 6x1 + 1 = o(c(x1))
    
    c(o(x1)) = 6x1 + 1 >= 6x1 + 1 = o(x1)
   problem:
    t(f(x1)) -> t(c(n(x1)))
    n(f(x1)) -> f(n(x1))
    n(s(x1)) -> f(s(x1))
    c(f(x1)) -> f(c(x1))
    c(n(x1)) -> n(c(x1))
    c(o(x1)) -> o(c(x1))
    c(o(x1)) -> o(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 1 1]     [0]
     [o](x0) = [0 0 0]x0 + [0]
               [0 0 1]     [1],

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