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TRS Standard pair #516964306
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
n066.star.cs.uiowa.edu
space
Beerendonk_07
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.441514015198 seconds
cpu usage
0.168308567
max memory
4812800.0
stage attributes
key
value
output-size
8579
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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))
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