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TRS Equational pair #487092702
details
property
value
status
complete
benchmark
AC16.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n148.star.cs.uiowa.edu
space
AProVE_AC_04
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.140945 seconds
cpu usage
0.157909
user time
0.077916
system time
0.079993
max virtual memory
113188.0
max residence set size
4704.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR x y) (THEORY (AC plus)) (RULES L(T(x)) -> L(x) L(plus(T(y),x)) -> plus(L(plus(x,y)),L(y)) T(T(x)) -> T(x) T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y)) plus(T(plus(x,y)),x) -> T(plus(x,y)) plus(T(x),x) -> T(x) plus(x,0) -> x plus(x,x) -> x ) Problem 1: Reduction Order Processor: -> Rules: L(T(x)) -> L(x) L(plus(T(y),x)) -> plus(L(plus(x,y)),L(y)) T(T(x)) -> T(x) T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y)) plus(T(plus(x,y)),x) -> T(plus(x,y)) plus(T(x),x) -> T(x) plus(x,0) -> x plus(x,x) -> x ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [L](X) = X + 1 [T](X) = 2.X + 2 [plus](X1,X2) = X1 + X2 + 1 [0] = 0 Problem 1: Reduction Order Processor: -> Rules: L(plus(T(y),x)) -> plus(L(plus(x,y)),L(y)) T(T(x)) -> T(x) T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y)) plus(T(plus(x,y)),x) -> T(plus(x,y)) plus(T(x),x) -> T(x) plus(x,0) -> x plus(x,x) -> x ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [L](X) = 2.X + 2 [T](X) = 2.X + 2 [plus](X1,X2) = X1 + X2 [0] = 0 Problem 1: Reduction Order Processor: -> Rules: T(T(x)) -> T(x) T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y)) plus(T(plus(x,y)),x) -> T(plus(x,y)) plus(T(x),x) -> T(x) plus(x,0) -> x plus(x,x) -> x ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [L](X) = 2.X [T](X) = 2.X + 2 [plus](X1,X2) = X1 + X2 [0] = 0 Problem 1: Reduction Order Processor: -> Rules:
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