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TRS Standard pair #516964547
details
property
value
status
complete
benchmark
perfect2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
0.254306077957 seconds
cpu usage
0.265102983
max memory
3.6528128E7
stage attributes
key
value
output-size
3523
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 = [f^#(s(_0),s(_1),_2,_3) -> f^#(s(_0),minus(_1,_0),_2,_3), f^#(s(_0),0,_1,_2) -> f^#(_0,_2,minus(_1,s(_0)),_2), f^#(s(_0),s(_1),_2,_3) -> f^#(_0,_3,_2,_3)] TRS = {minus(0,_0) -> 0, minus(s(_0),0) -> s(_0), minus(s(_0),s(_1)) -> minus(_0,_1), le(0,_0) -> true, le(s(_0),0) -> false, le(s(_0),s(_1)) -> le(_0,_1), if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, perfectp(0) -> false, perfectp(s(_0)) -> f(_0,s(0),s(_0),s(_0)), f(0,_0,0,_1) -> true, f(0,_0,s(_1),_2) -> false, f(s(_0),0,_1,_2) -> f(_0,_2,minus(_1,s(_0)),_2), f(s(_0),s(_1),_2,_3) -> if(le(_0,_1),f(s(_0),minus(_1,_0),_2,_3),f(_0,_3,_2,_3))} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## Trying with lexicographic path orders... The constraints are satisfied by the lexicographic path order using the argument filtering: {s:[0], perfectp:[0], le:[0, 1], if:[0, 1, 2], minus:[0], f:[0, 1, 2, 3], f^#:[0, 1, 2, 3]} and the precedence: s > [minus, 0, false, true], perfectp > [minus, s, if, le, 0, false, true, f], 0 > [true, false], f^# > [minus], f > [minus, if, le] This DP problem is finite. ## DP problem: Dependency pairs = [le^#(s(_0),s(_1)) -> le^#(_0,_1)] TRS = {minus(0,_0) -> 0, minus(s(_0),0) -> s(_0), minus(s(_0),s(_1)) -> minus(_0,_1), le(0,_0) -> true, le(s(_0),0) -> false, le(s(_0),s(_1)) -> le(_0,_1), if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, perfectp(0) -> false, perfectp(s(_0)) -> f(_0,s(0),s(_0),s(_0)), f(0,_0,0,_1) -> true, f(0,_0,s(_1),_2) -> false, f(s(_0),0,_1,_2) -> f(_0,_2,minus(_1,s(_0)),_2), f(s(_0),s(_1),_2,_3) -> if(le(_0,_1),f(s(_0),minus(_1,_0),_2,_3),f(_0,_3,_2,_3))} ## 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),0) -> s(_0), minus(s(_0),s(_1)) -> minus(_0,_1), le(0,_0) -> true, le(s(_0),0) -> false, le(s(_0),s(_1)) -> le(_0,_1), if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, perfectp(0) -> false, perfectp(s(_0)) -> f(_0,s(0),s(_0),s(_0)), f(0,_0,0,_1) -> true, f(0,_0,s(_1),_2) -> false, f(s(_0),0,_1,_2) -> f(_0,_2,minus(_1,s(_0)),_2), f(s(_0),s(_1),_2,_3) -> if(le(_0,_1),f(s(_0),minus(_1,_0),_2,_3),f(_0,_3,_2,_3))} ## 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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