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TRS Standard pair #516967117
details
property
value
status
complete
benchmark
#3.19.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n092.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
3.60100984573 seconds
cpu usage
3.786385218
max memory
1.91315968E8
stage attributes
key
value
output-size
3059
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 ** BEGIN proof argument ** All the DP problems were proved finite. As all the involved DP processors are sound, the TRS under analysis terminates. ** END proof argument ** ** BEGIN proof description ** ## Searching for a generalized rewrite rule (a rule whose right-hand side contains a variable that does not occur in the left-hand side)... No generalized rewrite rule found! ## Applying the DP framework... ## 3 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [3 DP problems]: ## DP problem: Dependency pairs = [plus^#(s(_0),_1) -> plus^#(_0,_1), plus^#(s(_0),_1) -> plus^#(_0,s(_1)), plus^#(s(_0),_1) -> plus^#(minus(_0,_1),double(_1)), plus^#(s(plus(_0,_1)),_2) -> plus^#(plus(_0,_1),_2), plus^#(s(plus(_0,_1)),_2) -> plus^#(_0,_1)] TRS = {minus(_0,0) -> _0, minus(s(_0),s(_1)) -> minus(_0,_1), double(0) -> 0, double(s(_0)) -> s(s(double(_0))), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), plus(s(_0),_1) -> plus(_0,s(_1)), plus(s(_0),_1) -> s(plus(minus(_0,_1),double(_1))), plus(s(plus(_0,_1)),_2) -> s(plus(plus(_0,_1),_2))} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Failed! ## Trying with lexicographic path orders... The constraints are satisfied by the lexicographic path order using the argument filtering: {double:[0], plus:[0, 1], s:[0], minus:[0], plus^#:[0, 1]} and the precedence: plus^# > [double, s, minus], double > [s, minus], plus > [plus^#, double, s, minus], s > [minus] This DP problem is finite. ## DP problem: Dependency pairs = [double^#(s(_0)) -> double^#(_0)] TRS = {minus(_0,0) -> _0, minus(s(_0),s(_1)) -> minus(_0,_1), double(0) -> 0, double(s(_0)) -> s(s(double(_0))), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), plus(s(_0),_1) -> plus(_0,s(_1)), plus(s(_0),_1) -> s(plus(minus(_0,_1),double(_1))), plus(s(plus(_0,_1)),_2) -> s(plus(plus(_0,_1),_2))} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [minus^#(s(_0),s(_1)) -> minus^#(_0,_1)] TRS = {minus(_0,0) -> _0, minus(s(_0),s(_1)) -> minus(_0,_1), double(0) -> 0, double(s(_0)) -> s(s(double(_0))), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), plus(s(_0),_1) -> plus(_0,s(_1)), plus(s(_0),_1) -> s(plus(minus(_0,_1),double(_1))), plus(s(plus(_0,_1)),_2) -> s(plus(plus(_0,_1),_2))} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## All the DP problems were proved finite. As all the involved DP processors are sound, the TRS under analysis terminates. Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64 using Java version 1.8.0_292 ** END proof description ** Total number of generated unfolded rules = 0
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