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TRS Stand 20472 pair #381711870
details
property
value
status
complete
benchmark
#4.35.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n051.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.712859153748 seconds
cpu usage
0.708297667
max memory
2.8823552E7
stage attributes
key
value
output-size
14982
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. and : [o * o] --> o apply : [o * o] --> o cons : [o * o] --> o eq : [o * o] --> o false : [] --> o if : [o * o * o] --> o lambda : [o * o] --> o nil : [] --> o ren : [o * o * o] --> o true : [] --> o var : [o] --> o and(true, X) => X and(false, X) => false eq(nil, nil) => true eq(cons(X, Y), nil) => false eq(nil, cons(X, Y)) => false eq(cons(X, Y), cons(Z, U)) => and(eq(X, Z), eq(Y, U)) eq(var(X), var(Y)) => eq(X, Y) eq(var(X), apply(Y, Z)) => false eq(var(X), lambda(Y, Z)) => false eq(apply(X, Y), var(Z)) => false eq(apply(X, Y), apply(Z, U)) => and(eq(X, Z), eq(Y, U)) eq(apply(X, Y), lambda(Z, X)) => false eq(lambda(X, Y), var(Z)) => false eq(lambda(X, Y), apply(Y, Z)) => false eq(lambda(X, Y), lambda(Z, U)) => and(eq(X, Z), eq(Y, U)) if(true, var(X), var(Y)) => var(X) if(false, var(X), var(Y)) => var(Y) ren(var(X), var(Y), var(Z)) => if(eq(X, Z), var(Y), var(Z)) ren(X, Y, apply(Z, U)) => apply(ren(X, Y, Z), ren(X, Y, U)) ren(X, Y, lambda(Z, U)) => lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), 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, all): Dependency Pairs P_0: 0] eq#(cons(X, Y), cons(Z, U)) =#> and#(eq(X, Z), eq(Y, U)) 1] eq#(cons(X, Y), cons(Z, U)) =#> eq#(X, Z) 2] eq#(cons(X, Y), cons(Z, U)) =#> eq#(Y, U) 3] eq#(var(X), var(Y)) =#> eq#(X, Y) 4] eq#(apply(X, Y), apply(Z, U)) =#> and#(eq(X, Z), eq(Y, U)) 5] eq#(apply(X, Y), apply(Z, U)) =#> eq#(X, Z) 6] eq#(apply(X, Y), apply(Z, U)) =#> eq#(Y, U) 7] eq#(lambda(X, Y), lambda(Z, U)) =#> and#(eq(X, Z), eq(Y, U)) 8] eq#(lambda(X, Y), lambda(Z, U)) =#> eq#(X, Z) 9] eq#(lambda(X, Y), lambda(Z, U)) =#> eq#(Y, U) 10] ren#(var(X), var(Y), var(Z)) =#> if#(eq(X, Z), var(Y), var(Z)) 11] ren#(var(X), var(Y), var(Z)) =#> eq#(X, Z) 12] ren#(X, Y, apply(Z, U)) =#> ren#(X, Y, Z) 13] ren#(X, Y, apply(Z, U)) =#> ren#(X, Y, U) 14] ren#(X, Y, lambda(Z, U)) =#> ren#(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U)) 15] ren#(X, Y, lambda(Z, U)) =#> ren#(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U) Rules R_0: and(true, X) => X and(false, X) => false eq(nil, nil) => true eq(cons(X, Y), nil) => false eq(nil, cons(X, Y)) => false eq(cons(X, Y), cons(Z, U)) => and(eq(X, Z), eq(Y, U)) eq(var(X), var(Y)) => eq(X, Y) eq(var(X), apply(Y, Z)) => false eq(var(X), lambda(Y, Z)) => false eq(apply(X, Y), var(Z)) => false eq(apply(X, Y), apply(Z, U)) => and(eq(X, Z), eq(Y, U)) eq(apply(X, Y), lambda(Z, X)) => false eq(lambda(X, Y), var(Z)) => false eq(lambda(X, Y), apply(Y, Z)) => false eq(lambda(X, Y), lambda(Z, U)) => and(eq(X, Z), eq(Y, U)) if(true, var(X), var(Y)) => var(X) if(false, var(X), var(Y)) => var(Y) ren(var(X), var(Y), var(Z)) => if(eq(X, Z), var(Y), var(Z)) ren(X, Y, apply(Z, U)) => apply(ren(X, Y, Z), ren(X, Y, U)) ren(X, Y, lambda(Z, U)) => lambda(var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), ren(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, U), nil)))), U))) Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite. We consider the dependency pair problem (P_0, R_0, minimal, all). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
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