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TRS Standard pair #487067092
details
property
value
status
complete
benchmark
pair3swap.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n180.star.cs.uiowa.edu
space
Endrullis_06
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
2.69858 seconds
cpu usage
9.2658
user time
7.96309
system time
1.30271
max virtual memory
5820680.0
max residence set size
111772.0
stage attributes
key
value
starexec-result
YES
output
YES Problem: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Proof: DP Processor: DPs: p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(x0)),x3) p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(b(x1))),p(a(a(x0)),x3)) p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) TRS: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Subterm Criterion Processor: simple projection: pi(p) = [1,1] pi(p#) = [1,1] problem: DPs: p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) TRS: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Matrix Interpretation Processor: dim=3 usable rules: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) interpretation: [0 0 0] [0 0 1] [p](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 1] [1 0 0] , [p#](x0, x1) = [1 0 1]x0 + [1 0 0]x1, [0 1 0] [0] [a](x0) = [0 0 0]x0 + [1] [1 0 1] [0], [0] [b](x0) = [0] [0] orientation: p#(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 1]x0 + [1 0 1]x2 + [1 0 0]x3 + [1] >= [1 1 1]x0 + [1 0 1]x2 + [1 0 0]x3 = p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) [0 0 0] [0 0 1] [0 0 0] [0 0 1] [0 0 0] [0 0 0] [0 0 1] p(a(a(x0)),p(x1,p(a(x2),x3))) = [0 0 0]x0 + [0 0 0]x1 + [0 0 0]x2 + [0 0 0]x3 >= [0 0 0]x0 + [0 0 0]x2 + [0 0 0]x3 = p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) [1 1 1] [0 0 0] [1 0 1] [1 0 0] [1 1 1] [0 0 1] [1 0 0] problem: DPs: TRS: p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) Qed
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