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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270381
details
property
value
status
complete
benchmark
h09.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Hamana_Kikuchi_18
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
1.11258 seconds
cpu usage
1.89654
user time
1.76707
system time
0.129476
max virtual memory
7029196.0
max residence set size
108600.0
stage attributes
key
value
starexec-result
YES
output
1.83/1.10 YES 1.83/1.11 We consider the system theBenchmark. 1.83/1.11 1.83/1.11 Alphabet: 1.83/1.11 1.83/1.11 0 : [] --> N 1.83/1.11 false : [] --> B 1.83/1.11 g : [] --> N -> B 1.83/1.11 h : [] --> N -> (N -> B) -> N -> B 1.83/1.11 iszero : [] --> N -> N -> B 1.83/1.11 rec : [] --> (N -> (N -> B) -> N -> B) -> B -> N -> B 1.83/1.11 true : [] --> B 1.83/1.11 1.83/1.11 Rules: 1.83/1.11 1.83/1.11 rec f (i 0) => i 1.83/1.11 g x => true 1.83/1.11 h x f y => false 1.83/1.11 iszero x y => rec h (g x) y 1.83/1.11 1.83/1.11 Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. 1.83/1.11 1.83/1.11 We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: 1.83/1.11 1.83/1.11 Alphabet: 1.83/1.11 1.83/1.11 0 : [] --> N 1.83/1.11 false : [] --> B 1.83/1.11 g : [] --> N -> B 1.83/1.11 h : [] --> N -> (N -> B) -> N -> B 1.83/1.11 iszero : [N] --> N -> B 1.83/1.11 rec : [N -> (N -> B) -> N -> B * B] --> N -> B 1.83/1.11 true : [] --> B 1.83/1.11 ~AP1 : [N -> B * N] --> B 1.83/1.11 1.83/1.11 Rules: 1.83/1.11 1.83/1.11 rec(F, ~AP1(G, 0)) => G 1.83/1.11 g X => true 1.83/1.11 h X F Y => false 1.83/1.11 iszero(X) Y => ~AP1(rec(h, g X), Y) 1.83/1.11 rec(F, g 0) => g 1.83/1.11 rec(F, h X G 0) => h X G 1.83/1.11 rec(F, iszero(X) 0) => iszero(X) 1.83/1.11 ~AP1(F, X) => F X 1.83/1.11 1.83/1.11 We observe that the rules contain a first-order subset: 1.83/1.11 1.83/1.11 g X => true 1.83/1.11 1.83/1.11 Moreover, the system is finitely branching. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS. 1.83/1.11 1.83/1.11 According to the external first-order termination prover, this system is indeed Ce-terminating: 1.83/1.11 1.83/1.11 || proof of resources/system.trs 1.83/1.11 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.83/1.11 || 1.83/1.11 || 1.83/1.11 || Termination w.r.t. Q of the given QTRS could be proven: 1.83/1.11 || 1.83/1.11 || (0) QTRS 1.83/1.11 || (1) QTRSRRRProof [EQUIVALENT] 1.83/1.11 || (2) QTRS 1.83/1.11 || (3) RisEmptyProof [EQUIVALENT] 1.83/1.11 || (4) YES 1.83/1.11 || 1.83/1.11 || 1.83/1.11 || ---------------------------------------- 1.83/1.11 || 1.83/1.11 || (0) 1.83/1.11 || Obligation: 1.83/1.11 || Q restricted rewrite system: 1.83/1.11 || The TRS R consists of the following rules: 1.83/1.11 || 1.83/1.11 || g(%X) -> true 1.83/1.11 || ~PAIR(%X, %Y) -> %X 1.83/1.11 || ~PAIR(%X, %Y) -> %Y 1.83/1.11 || 1.83/1.11 || Q is empty. 1.83/1.11 || 1.83/1.11 || ---------------------------------------- 1.83/1.11 || 1.83/1.11 || (1) QTRSRRRProof (EQUIVALENT) 1.83/1.11 || Used ordering: 1.83/1.11 || Polynomial interpretation [POLO]: 1.83/1.11 || 1.83/1.11 || POL(g(x_1)) = 2 + x_1 1.83/1.11 || POL(true) = 1 1.83/1.11 || POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2 1.83/1.11 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 1.83/1.11 || 1.83/1.11 || g(%X) -> true 1.83/1.11 || ~PAIR(%X, %Y) -> %X 1.83/1.11 || ~PAIR(%X, %Y) -> %Y 1.83/1.11 || 1.83/1.11 || 1.83/1.11 || 1.83/1.11 || 1.83/1.11 || ---------------------------------------- 1.83/1.11 ||
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