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TRS Standard pair #487068214
details
property
value
status
complete
benchmark
4.27.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n178.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.019691 seconds
cpu usage
0.020261
user time
0.010908
system time
0.009353
max virtual memory
113188.0
max residence set size
5324.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S x:S y:S) (RULES int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil ) Problem 1: Innermost Equivalent Processor: -> Rules: int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: INT(0,s(y:S)) -> INT(s(0),s(y:S)) INT(s(x:S),s(y:S)) -> INT(x:S,y:S) INT(s(x:S),s(y:S)) -> INT_LIST(int(x:S,y:S)) INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S) -> Rules: int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil Problem 1: SCC Processor: -> Pairs: INT(0,s(y:S)) -> INT(s(0),s(y:S)) INT(s(x:S),s(y:S)) -> INT(x:S,y:S) INT(s(x:S),s(y:S)) -> INT_LIST(int(x:S,y:S)) INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S) -> Rules: int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S) ->->-> Rules: int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil ->->Cycle: ->->-> Pairs: INT(0,s(y:S)) -> INT(s(0),s(y:S)) INT(s(x:S),s(y:S)) -> INT(x:S,y:S) ->->-> Rules: int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S) -> Rules: int(0,0) -> .(0,nil) int(0,s(y:S)) -> .(0,int(s(0),s(y:S))) int(s(x:S),0) -> nil int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S)) int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S)) int_list(nil) -> nil ->Projection: pi(INT_LIST) = 1 Problem 1.1:
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