Integer_Transition_Systems 2019-03-29 01.54 pair #432275856

details

property	value
status	complete
benchmark	ex32.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n078.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.81552 seconds
cpu usage	6.91323
user time	3.71151
system time	3.20171
max virtual memory	350452.0
max residence set size	8400.0

stage attributes

key	value
starexec-result	YES

output

6.85/6.81	YES
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
6.85/6.81	  f10#(I0, I1, I2, I3, I4) -> f4#(I0, 1, I2, I3, I4) 
6.85/6.81	  f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I5, I9) 
6.85/6.81	  f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	  f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	  f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	  f4#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4#(I50, I51, I52, I53, I54) -> f2#(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	The dependency graph for this problem is:
6.85/6.81	  0 -> 1
6.85/6.81	  1 -> 11
6.85/6.81	  2 -> 7
6.85/6.81	  3 -> 7
6.85/6.81	  4 -> 5
6.85/6.81	  5 -> 6
6.85/6.81	  6 -> 5
6.85/6.81	  7 -> 3, 4
6.85/6.81	  8 -> 12, 13
6.85/6.81	  9 -> 8
6.85/6.81	  10 -> 2
6.85/6.81	  11 -> 2
6.85/6.81	  12 -> 8
6.85/6.81	  13 -> 2
6.85/6.81	Where:
6.85/6.81	  0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
6.85/6.81	  1) f10#(I0, I1, I2, I3, I4) -> f4#(I0, 1, I2, I3, I4) 
6.85/6.81	  2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I5, I9) 
6.85/6.81	  3) f6#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  4) f6#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	  6) f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  7) f5#(I35, I36, I37, I38, I39) -> f6#(I35, I36, I37, I38, I39) 
6.85/6.81	  8) f3#(I40, I41, I42, I43, I44) -> f1#(I40, I41, I42, I43, I44) 
6.85/6.81	  9) f4#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  10) f4#(I50, I51, I52, I53, I54) -> f2#(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  11) f4#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  12) f1#(I60, I61, I62, I63, I64) -> f3#(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  13) f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	We have the following SCCs.
6.85/6.81	  { 8, 12 }
6.85/6.81	  { 3, 7 }
6.85/6.81	  { 5, 6 }
6.85/6.81	
6.85/6.81	DP problem for innermost termination.
6.85/6.81	P =
6.85/6.81	  f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) 
6.85/6.81	  f7#(I25, I26, I27, I28, I29) -> f9#(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	R = 
6.85/6.81	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
6.85/6.81	  f10(I0, I1, I2, I3, I4) -> f4(I0, 1, I2, I3, I4) 
6.85/6.81	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I5, I9) 
6.85/6.81	  f6(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, -1 + I13, I14) [0 <= I13]
6.85/6.81	  f6(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I15) [1 + I18 <= 0]
6.85/6.81	  f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) 
6.85/6.81	  f7(I25, I26, I27, I28, I29) -> f9(I25, I26, I27, I28, -1 + I29) [0 <= I29]
6.85/6.81	  f7(I30, I31, I32, I33, I34) -> f8(I30, I31, I32, I33, I34) [1 + I34 <= 0]
6.85/6.81	  f5(I35, I36, I37, I38, I39) -> f6(I35, I36, I37, I38, I39) 
6.85/6.81	  f3(I40, I41, I42, I43, I44) -> f1(I40, I41, I42, I43, I44) 
6.85/6.81	  f4(I45, I46, I47, I48, I49) -> f3(I45, I46, I45, I48, I49) [0 <= I46 /\ I46 <= 0]
6.85/6.81	  f4(I50, I51, I52, I53, I54) -> f2(I50, I51, I52, I53, I54) [1 + I51 <= 0]
6.85/6.81	  f4(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, I58, I59) [1 <= I56]
6.85/6.81	  f1(I60, I61, I62, I63, I64) -> f3(I60, I61, -1 + I62, I63, I64) [0 <= I62]
6.85/6.81	  f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I67 <= 0]
6.85/6.81	
6.85/6.81	We use the basic value criterion with the projection function NU:
6.85/6.81	  NU[f7#(z1,z2,z3,z4,z5)] = z5
6.85/6.81	  NU[f9#(z1,z2,z3,z4,z5)] = z5
6.85/6.81	
6.85/6.81	This gives the following inequalities:
6.85/6.81	    ==>  I24 (>! \union =) I24
6.85/6.81	  0 <= I29  ==>  I29 >! -1 + I29
6.85/6.81

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