details

property	value
status	complete
benchmark	AC07.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	AProVE_AC_04

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	0.800863 seconds
cpu usage	0.722173
user time	0.397856
system time	0.324317
max virtual memory	434836.0
max residence set size	6004.0

stage attributes

key	value
starexec-result	YES

output

YES Problem 1: (VAR x xs y) (THEORY (AC plus)) (RULES S(cons(x,xs)) -> plus(x,S(xs)) S(nil) -> 0 int(0,0) -> cons(0,nil) int(0,s(y)) -> cons(0,int(s(0),s(y))) int(s(x),0) -> nil int(s(x),s(y)) -> intlist(int(x,y)) intlist(cons(x,y)) -> cons(s(x),intlist(y)) intlist(nil) -> nil plus(x,0) -> x plus(x,s(y)) -> s(plus(x,y)) sum(x,y) -> S(int(x,y)) ) Problem 1: Reduction Order Processor: -> Rules: S(cons(x,xs)) -> plus(x,S(xs)) S(nil) -> 0 int(0,0) -> cons(0,nil) int(0,s(y)) -> cons(0,int(s(0),s(y))) int(s(x),0) -> nil int(s(x),s(y)) -> intlist(int(x,y)) intlist(cons(x,y)) -> cons(s(x),intlist(y)) intlist(nil) -> nil plus(x,0) -> x plus(x,s(y)) -> s(plus(x,y)) sum(x,y) -> S(int(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [S](X) = X [int](X1,X2) = 2.X1 + 2.X2 + 2 [intlist](X) = X [plus](X1,X2) = X1 + X2 [sum](X1,X2) = 2.X1 + 2.X2 + 2 [0] = 0 [cons](X1,X2) = 2.X1 + X2 [nil] = 2 [s](X) = X Problem 1: Reduction Order Processor: -> Rules: S(cons(x,xs)) -> plus(x,S(xs)) int(0,0) -> cons(0,nil) int(0,s(y)) -> cons(0,int(s(0),s(y))) int(s(x),0) -> nil int(s(x),s(y)) -> intlist(int(x,y)) intlist(cons(x,y)) -> cons(s(x),intlist(y)) intlist(nil) -> nil plus(x,0) -> x plus(x,s(y)) -> s(plus(x,y)) sum(x,y) -> S(int(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [S](X) = X + 1 [int](X1,X2) = 2.X1 + 2.X2 + 1 [intlist](X) = X [plus](X1,X2) = X1 + X2 [sum](X1,X2) = 2.X1 + 2.X2 + 2 [0] = 0 [cons](X1,X2) = 2.X1 + X2 [nil] = 0 [s](X) = X Problem 1: Reduction Order Processor: -> Rules: S(cons(x,xs)) -> plus(x,S(xs)) int(0,s(y)) -> cons(0,int(s(0),s(y))) int(s(x),0) -> nil int(s(x),s(y)) -> intlist(int(x,y)) intlist(cons(x,y)) -> cons(s(x),intlist(y)) intlist(nil) -> nil popout

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

job log

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actions

all output return to TRS Equational