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TRS Standard pair #487074024
details
property
value
status
complete
benchmark
append-hard.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n191.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.135065 seconds
cpu usage
0.073219
user time
0.034966
system time
0.038253
max virtual memory
113188.0
max residence set size
5472.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S l:S l1:S l2:S x:S) (RULES append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S)) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S)) ifappend(l1:S,l2:S,ttrue) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S ) Problem 1: Innermost Equivalent Processor: -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S)) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S)) ifappend(l1:S,l2:S,ttrue) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S)) APPEND(l1:S,l2:S) -> IS_EMPTY(l1:S) IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S) IFAPPEND(l1:S,l2:S,ffalse) -> HD(l1:S) IFAPPEND(l1:S,l2:S,ffalse) -> TL(l1:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S)) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S)) ifappend(l1:S,l2:S,ttrue) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S Problem 1: SCC Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S)) APPEND(l1:S,l2:S) -> IS_EMPTY(l1:S) IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S) IFAPPEND(l1:S,l2:S,ffalse) -> HD(l1:S) IFAPPEND(l1:S,l2:S,ffalse) -> TL(l1:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S)) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S)) ifappend(l1:S,l2:S,ttrue) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S)) IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S) ->->-> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S)) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S)) ifappend(l1:S,l2:S,ttrue) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S Problem 1: Reduction Pairs Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S)) IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S)) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S)) ifappend(l1:S,l2:S,ttrue) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S -> Usable rules: is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S ->Interpretation type: Linear ->Coefficients: All rationals
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