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TRS Equational pair #487092813
details
property
value
status
complete
benchmark
YWHM14_3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Mixed_AC
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.258807 seconds
cpu usage
0.27788
user time
0.153476
system time
0.124404
max virtual memory
113188.0
max residence set size
5220.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR x y) (THEORY (AC plus)) (RULES f(plus(x,y)) -> plus(f(x),y) h(a,b) -> h(b,a) h(a,g(g(a))) -> h(g(a),f(a)) h(g(a),a) -> h(a,g(b)) h(g(a),b) -> h(a,g(a)) plus(f(a),g(b)) -> plus(f(b),g(a)) plus(g(x),y) -> g(plus(x,y)) ) Problem 1: Reduction Order Processor: -> Rules: f(plus(x,y)) -> plus(f(x),y) h(a,b) -> h(b,a) h(a,g(g(a))) -> h(g(a),f(a)) h(g(a),a) -> h(a,g(b)) h(g(a),b) -> h(a,g(a)) plus(f(a),g(b)) -> plus(f(b),g(a)) plus(g(x),y) -> g(plus(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [f](X) = 2.X [h](X1,X2) = 2.X1 + 2.X2 [plus](X1,X2) = X1 + X2 + 1 [a] = 1 [b] = 1 [g](X) = X + 2 Problem 1: Reduction Order Processor: -> Rules: h(a,b) -> h(b,a) h(a,g(g(a))) -> h(g(a),f(a)) h(g(a),a) -> h(a,g(b)) h(g(a),b) -> h(a,g(a)) plus(f(a),g(b)) -> plus(f(b),g(a)) plus(g(x),y) -> g(plus(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [f](X) = X [h](X1,X2) = 2.X1 + X2 [plus](X1,X2) = X1 + X2 + 1 [a] = 2 [b] = 1 [g](X) = X + 2 Problem 1: Reduction Order Processor: -> Rules: h(a,g(g(a))) -> h(g(a),f(a)) h(g(a),a) -> h(a,g(b)) h(g(a),b) -> h(a,g(a)) plus(f(a),g(b)) -> plus(f(b),g(a)) plus(g(x),y) -> g(plus(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [f](X) = 2.X [h](X1,X2) = 2.X1 + 2.X2 [plus](X1,X2) = X1 + X2 [a] = 1 [b] = 1 [g](X) = X + 2 Problem 1:
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