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TRS Standard pair #487072634
details
property
value
status
complete
benchmark
10.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
Beerendonk_07
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.221035 seconds
cpu usage
0.168053
user time
0.108769
system time
0.059284
max virtual memory
113188.0
max residence set size
5840.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S x:S) (RULES cond1(ttrue,x:S) -> cond2(even(x:S),x:S) cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S)) cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S)) div2(0) -> 0 div2(s(0)) -> 0 div2(s(s(x:S))) -> s(div2(x:S)) even(0) -> ttrue even(s(0)) -> ffalse even(s(s(x:S))) -> even(x:S) neq(0,0) -> ffalse neq(0,s(x:S)) -> ttrue neq(s(x:S),0) -> ttrue neq(s(x:S),s(y)) -> neq(x:S,y) p(0) -> 0 p(s(x:S)) -> x:S ) Problem 1: Innermost Equivalent Processor: -> Rules: cond1(ttrue,x:S) -> cond2(even(x:S),x:S) cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S)) cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S)) div2(0) -> 0 div2(s(0)) -> 0 div2(s(s(x:S))) -> s(div2(x:S)) even(0) -> ttrue even(s(0)) -> ffalse even(s(s(x:S))) -> even(x:S) neq(0,0) -> ffalse neq(0,s(x:S)) -> ttrue neq(s(x:S),0) -> ttrue neq(s(x:S),s(y)) -> neq(x:S,y) p(0) -> 0 p(s(x:S)) -> x:S -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: COND1(ttrue,x:S) -> COND2(even(x:S),x:S) COND1(ttrue,x:S) -> EVEN(x:S) COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S)) COND2(ffalse,x:S) -> NEQ(x:S,0) COND2(ffalse,x:S) -> P(x:S) COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S)) COND2(ttrue,x:S) -> DIV2(x:S) COND2(ttrue,x:S) -> NEQ(x:S,0) DIV2(s(s(x:S))) -> DIV2(x:S) EVEN(s(s(x:S))) -> EVEN(x:S) -> Rules: cond1(ttrue,x:S) -> cond2(even(x:S),x:S) cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S)) cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S)) div2(0) -> 0 div2(s(0)) -> 0 div2(s(s(x:S))) -> s(div2(x:S)) even(0) -> ttrue even(s(0)) -> ffalse even(s(s(x:S))) -> even(x:S) neq(0,0) -> ffalse neq(0,s(x:S)) -> ttrue neq(s(x:S),0) -> ttrue neq(s(x:S),s(y)) -> neq(x:S,y) p(0) -> 0 p(s(x:S)) -> x:S Problem 1: SCC Processor: -> Pairs: COND1(ttrue,x:S) -> COND2(even(x:S),x:S) COND1(ttrue,x:S) -> EVEN(x:S) COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S)) COND2(ffalse,x:S) -> NEQ(x:S,0) COND2(ffalse,x:S) -> P(x:S) COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S)) COND2(ttrue,x:S) -> DIV2(x:S) COND2(ttrue,x:S) -> NEQ(x:S,0) DIV2(s(s(x:S))) -> DIV2(x:S) EVEN(s(s(x:S))) -> EVEN(x:S) -> Rules: cond1(ttrue,x:S) -> cond2(even(x:S),x:S) cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S)) cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S)) div2(0) -> 0 div2(s(0)) -> 0 div2(s(s(x:S))) -> s(div2(x:S)) even(0) -> ttrue even(s(0)) -> ffalse even(s(s(x:S))) -> even(x:S)
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