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TRS Standard pair #487067587
details
property
value
status
complete
benchmark
#3.2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n175.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.653716 seconds
cpu usage
1.47284
user time
1.0717
system time
0.401139
max virtual memory
3981440.0
max residence set size
66436.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(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#(x,s(y)) -> minus#(x,y) minus#(x,s(y)) -> pred#(minus(x,y)) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(minus(x,y)) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) TDG Processor: DPs: minus#(x,s(y)) -> minus#(x,y) minus#(x,s(y)) -> pred#(minus(x,y)) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(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#(x,s(y)) -> pred#(minus(x,y)) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> minus#(x,y) minus#(x,s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> pred#(minus(x,y)) minus#(x,s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> minus#(x,y) SCC Processor: #sccs: 2 #rules: 2 #arcs: 6/16 DPs: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(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(s) = [0,0,0] pi(pred) = 0 pi(minus) = [0,0] pi(quot#) = 0 problem: DPs: TRS: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(minus(x,y)) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Qed DPs: minus#(x,s(y)) -> minus#(x,y) TRS: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(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#) = 1 problem: DPs: TRS: pred(s(x)) -> x minus(x,0()) -> x minus(x,s(y)) -> pred(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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