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TRS Standard pair #487067442
details
property
value
status
complete
benchmark
#3.1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n179.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.500504 seconds
cpu usage
0.899904
user time
0.668834
system time
0.23107
max virtual memory
96176.0
max residence set size
63696.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Proof: DP Processor: DPs: minus#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) TDG Processor: DPs: minus#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) graph: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) -> quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) -> quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(s(x),s(y)) -> minus#(x,y) minus#(s(x),s(y)) -> minus#(x,y) -> minus#(s(x),s(y)) -> minus#(x,y) SCC Processor: #sccs: 2 #rules: 2 #arcs: 4/9 DPs: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Subterm Criterion Processor: simple projection: pi(minus) = [0,0] pi(s) = [0,0,0] pi(quot#) = 0 problem: DPs: TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Qed DPs: minus#(s(x),s(y)) -> minus#(x,y) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Subterm Criterion Processor: simple projection: pi(minus#) = 0 problem: DPs: TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Qed
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