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TRS Conditional pair #487562997
details
property
value
status
complete
benchmark
logic.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n147.star.cs.uiowa.edu
space
Mixed_CTRS_2014
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
117.650244951 seconds
cpu usage
117.393160641
max memory
1.06827776E8
stage attributes
key
value
output-size
10901
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR v_NonEmpty:S x:S y:S) (RULES and(not(x:S),x:S) -> 0 and(0,x:S) -> 0 and(1,x:S) -> x:S and(x:S,not(x:S)) -> 0 and(x:S,0) -> 0 and(x:S,1) -> x:S f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1 implies(x:S,y:S) -> 0 | x:S ->* 1, y:S ->* 0 implies(x:S,y:S) -> 1 | not(x:S) ->* 1 implies(x:S,y:S) -> 1 | y:S ->* 1 not(0) -> 1 not(1) -> 0 or(not(x:S),x:S) -> 1 or(0,x:S) -> x:S or(1,x:S) -> 1 or(x:S,not(x:S)) -> 1 or(x:S,0) -> x:S or(x:S,1) -> 1 ) Problem 1: Valid CTRS Processor: -> Rules: and(not(x:S),x:S) -> 0 and(0,x:S) -> 0 and(1,x:S) -> x:S and(x:S,not(x:S)) -> 0 and(x:S,0) -> 0 and(x:S,1) -> x:S f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1 implies(x:S,y:S) -> 0 | x:S ->* 1, y:S ->* 0 implies(x:S,y:S) -> 1 | not(x:S) ->* 1 implies(x:S,y:S) -> 1 | y:S ->* 1 not(0) -> 1 not(1) -> 0 or(not(x:S),x:S) -> 1 or(0,x:S) -> x:S or(1,x:S) -> 1 or(x:S,not(x:S)) -> 1 or(x:S,0) -> x:S or(x:S,1) -> 1 -> The system is a deterministic 3-CTRS. Problem 1: Dependency Pairs Processor: Conditional Termination Problem 1: -> Pairs: F(x:S) -> F(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1 -> QPairs: Empty -> Rules: and(not(x:S),x:S) -> 0 and(0,x:S) -> 0 and(1,x:S) -> x:S and(x:S,not(x:S)) -> 0 and(x:S,0) -> 0 and(x:S,1) -> x:S f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1 implies(x:S,y:S) -> 0 | x:S ->* 1, y:S ->* 0 implies(x:S,y:S) -> 1 | not(x:S) ->* 1 implies(x:S,y:S) -> 1 | y:S ->* 1 not(0) -> 1 not(1) -> 0 or(not(x:S),x:S) -> 1 or(0,x:S) -> x:S or(1,x:S) -> 1 or(x:S,not(x:S)) -> 1 or(x:S,0) -> x:S or(x:S,1) -> 1 Conditional Termination Problem 2: -> Pairs: F(x:S) -> IMPLIES(implies(x:S,implies(x:S,0)),0) F(x:S) -> IMPLIES(x:S,implies(x:S,0)) F(x:S) -> IMPLIES(x:S,0) IMPLIES(x:S,y:S) -> NOT(x:S) -> QPairs: Empty -> Rules: and(not(x:S),x:S) -> 0 and(0,x:S) -> 0 and(1,x:S) -> x:S and(x:S,not(x:S)) -> 0 and(x:S,0) -> 0 and(x:S,1) -> x:S f(x:S) -> f(0) | implies(implies(x:S,implies(x:S,0)),0) ->* 1
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