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TRS Stand 20472 pair #381714364
details
property
value
status
complete
benchmark
kabasci04.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n070.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.157642841339 seconds
cpu usage
0.154226364
max memory
5918720.0
stage attributes
key
value
output-size
8855
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o any : [o] --> o gcd : [o * o] --> o max : [o * o] --> o min : [o * o] --> o minus : [o * o] --> o s : [o] --> o min(X, 0) => 0 min(0, X) => 0 min(s(X), s(Y)) => s(min(X, Y)) max(X, 0) => X max(0, X) => X max(s(X), s(Y)) => s(max(X, Y)) minus(X, 0) => X minus(s(X), s(Y)) => s(minus(X, any(Y))) gcd(s(X), s(Y)) => gcd(minus(max(X, Y), min(X, Y)), s(min(X, Y))) any(s(X)) => s(s(any(X))) any(X) => X We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] min#(s(X), s(Y)) =#> min#(X, Y) 1] max#(s(X), s(Y)) =#> max#(X, Y) 2] minus#(s(X), s(Y)) =#> minus#(X, any(Y)) 3] minus#(s(X), s(Y)) =#> any#(Y) 4] gcd#(s(X), s(Y)) =#> gcd#(minus(max(X, Y), min(X, Y)), s(min(X, Y))) 5] gcd#(s(X), s(Y)) =#> minus#(max(X, Y), min(X, Y)) 6] gcd#(s(X), s(Y)) =#> max#(X, Y) 7] gcd#(s(X), s(Y)) =#> min#(X, Y) 8] gcd#(s(X), s(Y)) =#> min#(X, Y) 9] any#(s(X)) =#> any#(X) Rules R_0: min(X, 0) => 0 min(0, X) => 0 min(s(X), s(Y)) => s(min(X, Y)) max(X, 0) => X max(0, X) => X max(s(X), s(Y)) => s(max(X, Y)) minus(X, 0) => X minus(s(X), s(Y)) => s(minus(X, any(Y))) gcd(s(X), s(Y)) => gcd(minus(max(X, Y), min(X, Y)), s(min(X, Y))) any(s(X)) => s(s(any(X))) any(X) => X Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 1 * 2 : 2, 3 * 3 : 9 * 4 : 4, 5, 6, 7, 8 * 5 : 2, 3 * 6 : 1 * 7 : 0 * 8 : 0 * 9 : 9 This graph has the following strongly connected components: P_1: min#(s(X), s(Y)) =#> min#(X, Y) P_2: max#(s(X), s(Y)) =#> max#(X, Y) P_3: minus#(s(X), s(Y)) =#> minus#(X, any(Y)) P_4: gcd#(s(X), s(Y)) =#> gcd#(minus(max(X, Y), min(X, Y)), s(min(X, Y))) P_5: any#(s(X)) =#> any#(X)
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