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TRS Stand 20472 pair #381711994
details
property
value
status
complete
benchmark
#3.57.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n113.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.315140008926 seconds
cpu usage
0.301875192
max memory
1.1001856E7
stage attributes
key
value
output-size
12751
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o app : [o * o] --> o cons : [o * o] --> o minus : [o * o] --> o nil : [] --> o plus : [o * o] --> o quot : [o * o] --> o s : [o] --> o sum : [o] --> o minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) 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] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 2] quot#(s(X), s(Y)) =#> minus#(X, Y) 3] plus#(s(X), Y) =#> plus#(X, Y) 4] minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z)) 5] minus#(minus(X, Y), Z) =#> plus#(Y, Z) 6] app#(cons(X, Y), Z) =#> app#(Y, Z) 7] sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z)) 8] sum#(cons(X, cons(Y, Z))) =#> plus#(X, Y) 9] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U))))) 10] sum#(app(X, cons(Y, cons(Z, U)))) =#> app#(X, sum(cons(Y, cons(Z, U)))) 11] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(cons(Y, cons(Z, U))) Rules R_0: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) 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, 4, 5 * 1 : 1, 2 * 2 : 0, 4, 5 * 3 : 3 * 4 : 0, 4, 5 * 5 : 3 * 6 : 6 * 7 : 7, 8 * 8 : 3 * 9 : 7, 8, 9, 10, 11 * 10 : 6 * 11 : 7, 8 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z)) P_2: quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y))
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