Integer_Transition_Systems 2019-03-29 01.54 pair #432274026

details
property	value
status	complete
benchmark	constants.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n182.star.cs.uiowa.edu
space	From_T2
run statistics
property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	10.904 seconds
cpu usage	11.092
user time	5.48148
system time	5.61054
max virtual memory	697316.0
max residence set size	11200.0
stage attributes
key	value
starexec-result	YES
output
11.03/10.90	YES
11.03/10.90	
11.03/10.90	DP problem for innermost termination.
11.03/10.90	P =
11.03/10.90	  f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f5#(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4#(I44, I45, I46, I47, I48, I49, I50) -> f5#(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3#(I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	R = 
11.03/10.90	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0]
11.03/10.90	  f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34]
11.03/10.90	  f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	The dependency graph for this problem is:
11.03/10.90	  0 -> 1
11.03/10.90	  1 -> 3
11.03/10.90	  2 -> 3, 4, 5
11.03/10.90	  3 -> 2
11.03/10.90	  4 -> 
11.03/10.90	  5 -> 
11.03/10.90	  6 -> 3, 4, 5
11.03/10.90	  7 -> 2
11.03/10.90	Where:
11.03/10.90	  0) f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  1) f8#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  3) f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  4) f4#(I37, I38, I39, I40, I41, I42, I43) -> f5#(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  5) f4#(I44, I45, I46, I47, I48, I49, I50) -> f5#(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  6) f3#(I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  7) f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	We have the following SCCs.
11.03/10.90	  { 2, 3 }
11.03/10.90	
11.03/10.90	DP problem for innermost termination.
11.03/10.90	P =
11.03/10.90	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	R = 
11.03/10.90	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0]
11.03/10.90	  f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34]
11.03/10.90	  f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	We use the reverse value criterion with the projection function NU:
11.03/10.90	  NU[f4#(z1,z2,z3,z4,z5,z6,z7)] = -1 + z4 - z5 + -1 * 0
11.03/10.90	  NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = -1 + z4 - (1 + z5) + -1 * 0
11.03/10.90	
11.03/10.90	This gives the following inequalities:
11.03/10.90	    ==>  -1 + I10 - (1 + I11) + -1 * 0 >= -1 + I10 - (1 + I11) + -1 * 0
11.03/10.90	  I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1  ==>  -1 + I17 - I18 + -1 * 0 > -1 + I17 - (1 + I18) + -1 * 0  with  -1 + I17 - I18 + -1 * 0 >= 0
11.03/10.90	
11.03/10.90	We remove all the strictly oriented dependency pairs.
11.03/10.90	
11.03/10.90	DP problem for innermost termination.
11.03/10.90	P =
11.03/10.90	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	R = 
11.03/10.90	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0]
11.03/10.90	  f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34]
11.03/10.90	  f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	The dependency graph for this problem is:
11.03/10.90	  2 -> 
11.03/10.90	Where:
11.03/10.90	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	
11.03/10.90	We have the following SCCs.
11.03/10.90	  
11.03/13.87	EOF
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