TRS Condi 20667 pair #381733078

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	n085.star.cs.uiowa.edu
space	Mixed_CTRS_2014

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	12.0366568565 seconds
cpu usage	11.110354714
max memory	6.9148672E7

stage attributes

key	value
output-size	8632
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 x y)
(RULES
and(not(x),x) -> 0
and(0,x) -> 0
and(1,x) -> x
and(x,not(x)) -> 0
and(x,0) -> 0
and(x,1) -> x
f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
implies(x,y) -> 0 | x -> 1, y -> 0
implies(x,y) -> 1 | not(x) -> 1
implies(x,y) -> 1 | y -> 1
not(0) -> 1
not(1) -> 0
or(not(x),x) -> 1
or(0,x) -> x
or(1,x) -> 1
or(x,not(x)) -> 1
or(x,0) -> x
or(x,1) -> 1
)

Problem 1: 
Valid CTRS Processor:
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1
-> The system is a deterministic 3-CTRS.

Problem 1: 

Dependency Pairs Processor:

Conditional Termination Problem 1:
-> Pairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> QPairs:
 Empty
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1
 implies(x,y) -> 0 | x -> 1, y -> 0
 implies(x,y) -> 1 | not(x) -> 1
 implies(x,y) -> 1 | y -> 1
 not(0) -> 1
 not(1) -> 0
 or(not(x),x) -> 1
 or(0,x) -> x
 or(1,x) -> 1
 or(x,not(x)) -> 1
 or(x,0) -> x
 or(x,1) -> 1

Conditional Termination Problem 2:
-> Pairs:
 F(x) -> IMPLIES(implies(x,implies(x,0)),0)
 F(x) -> IMPLIES(x,implies(x,0))
 F(x) -> IMPLIES(x,0)
 IMPLIES(x,y) -> NOT(x)
-> QPairs:
 F(x) -> F(0) | implies(implies(x,implies(x,0)),0) -> 1
-> Rules:
 and(not(x),x) -> 0
 and(0,x) -> 0
 and(1,x) -> x
 and(x,not(x)) -> 0
 and(x,0) -> 0
 and(x,1) -> x
 f(x) -> f(0) | implies(implies(x,implies(x,0)),0) -> 1

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