SRS Standard pair #487089374

details

property	value
status	complete
benchmark	aprove5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Secret_05_SRS

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	47.2287 seconds
cpu usage	184.769
user time	178.232
system time	6.5375
max virtual memory	1.1688912E7
max residence set size	151316.0

stage attributes

key	value
starexec-result	YES

output

YES

Problem:
 a(b(x1)) -> b(c(a(x1)))
 b(c(x1)) -> c(b(b(x1)))
 a(c(x1)) -> c(a(b(x1)))
 a(a(x1)) -> a(d(d(d(x1))))
 d(a(x1)) -> d(d(c(x1)))
 a(d(d(c(x1)))) -> a(a(a(d(x1))))
 e(e(f(f(x1)))) -> f(f(f(e(e(x1)))))
 e(x1) -> a(x1)
 b(d(x1)) -> d(d(x1))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [f](x0) = x0,
   
   [a](x0) = x0,
   
   [d](x0) = x0,
   
   [e](x0) = x0 + 4,
   
   [b](x0) = x0,
   
   [c](x0) = x0
  orientation:
   a(b(x1)) = x1 >= x1 = b(c(a(x1)))
   
   b(c(x1)) = x1 >= x1 = c(b(b(x1)))
   
   a(c(x1)) = x1 >= x1 = c(a(b(x1)))
   
   a(a(x1)) = x1 >= x1 = a(d(d(d(x1))))
   
   d(a(x1)) = x1 >= x1 = d(d(c(x1)))
   
   a(d(d(c(x1)))) = x1 >= x1 = a(a(a(d(x1))))
   
   e(e(f(f(x1)))) = x1 + 8 >= x1 + 8 = f(f(f(e(e(x1)))))
   
   e(x1) = x1 + 4 >= x1 = a(x1)
   
   b(d(x1)) = x1 >= x1 = d(d(x1))
  problem:
   a(b(x1)) -> b(c(a(x1)))
   b(c(x1)) -> c(b(b(x1)))
   a(c(x1)) -> c(a(b(x1)))
   a(a(x1)) -> a(d(d(d(x1))))
   d(a(x1)) -> d(d(c(x1)))
   a(d(d(c(x1)))) -> a(a(a(d(x1))))
   e(e(f(f(x1)))) -> f(f(f(e(e(x1)))))
   b(d(x1)) -> d(d(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0) = x0 + 1,
    
    [a](x0) = x0,
    
    [d](x0) = x0,
    
    [e](x0) = 2x0 + 3,
    
    [b](x0) = x0,
    
    [c](x0) = x0
   orientation:
    a(b(x1)) = x1 >= x1 = b(c(a(x1)))
    
    b(c(x1)) = x1 >= x1 = c(b(b(x1)))
    
    a(c(x1)) = x1 >= x1 = c(a(b(x1)))
    
    a(a(x1)) = x1 >= x1 = a(d(d(d(x1))))
    
    d(a(x1)) = x1 >= x1 = d(d(c(x1)))
    
    a(d(d(c(x1)))) = x1 >= x1 = a(a(a(d(x1))))
    
    e(e(f(f(x1)))) = 4x1 + 17 >= 4x1 + 12 = f(f(f(e(e(x1)))))
    
    b(d(x1)) = x1 >= x1 = d(d(x1))
   problem:
    a(b(x1)) -> b(c(a(x1)))
    b(c(x1)) -> c(b(b(x1)))
    a(c(x1)) -> c(a(b(x1)))
    a(a(x1)) -> a(d(d(d(x1))))
    d(a(x1)) -> d(d(c(x1)))
    a(d(d(c(x1)))) -> a(a(a(d(x1))))
    b(d(x1)) -> d(d(x1))
   String Reversal Processor:
    b(a(x1)) -> a(c(b(x1)))
    c(b(x1)) -> b(b(c(x1)))
    c(a(x1)) -> b(a(c(x1)))
    a(a(x1)) -> d(d(d(a(x1))))
    a(d(x1)) -> c(d(d(x1)))
    c(d(d(a(x1)))) -> d(a(a(a(x1))))

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job log

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