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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270661
details
property
value
status
complete
benchmark
Applicative_first_order_05__#3.25.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n134.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.829006 seconds
cpu usage
1.70606
user time
1.57623
system time
0.129829
max virtual memory
6817872.0
max residence set size
125080.0
stage attributes
key
value
starexec-result
YES
output
1.62/0.82 YES 1.62/0.82 We consider the system theBenchmark. 1.62/0.82 1.62/0.82 Alphabet: 1.62/0.82 1.62/0.82 cons : [c * d] --> d 1.62/0.82 f : [a] --> a 1.62/0.82 false : [] --> b 1.62/0.82 filter : [c -> b * d] --> d 1.62/0.82 filter2 : [b * c -> b * c * d] --> d 1.62/0.82 g : [a] --> a 1.62/0.82 h : [a] --> a 1.62/0.82 map : [c -> c * d] --> d 1.62/0.82 nil : [] --> d 1.62/0.82 true : [] --> b 1.62/0.82 1.62/0.82 Rules: 1.62/0.82 1.62/0.82 f(g(x)) => g(f(f(x))) 1.62/0.82 f(h(x)) => h(g(x)) 1.62/0.82 map(i, nil) => nil 1.62/0.82 map(i, cons(x, y)) => cons(i x, map(i, y)) 1.62/0.82 filter(i, nil) => nil 1.62/0.82 filter(i, cons(x, y)) => filter2(i x, i, x, y) 1.62/0.82 filter2(true, i, x, y) => cons(x, filter(i, y)) 1.62/0.82 filter2(false, i, x, y) => filter(i, y) 1.62/0.82 1.62/0.82 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.62/0.82 1.62/0.82 We use rule removal, following [Kop12, Theorem 2.23]. 1.62/0.82 1.62/0.82 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.62/0.82 1.62/0.82 f(g(X)) >? g(f(f(X))) 1.62/0.82 f(h(X)) >? h(g(X)) 1.62/0.82 map(F, nil) >? nil 1.62/0.82 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 1.62/0.82 filter(F, nil) >? nil 1.62/0.82 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 1.62/0.82 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 1.62/0.82 filter2(false, F, X, Y) >? filter(F, Y) 1.62/0.82 1.62/0.82 We orient these requirements with a polynomial interpretation in the natural numbers. 1.62/0.82 1.62/0.82 The following interpretation satisfies the requirements: 1.62/0.82 1.62/0.82 cons = \y0y1.1 + y0 + y1 1.62/0.82 f = \y0.y0 1.62/0.82 false = 3 1.62/0.82 filter = \G0y1.2 + 2y1 + G0(0) + 2y1G0(y1) 1.62/0.82 filter2 = \y0G1y2y3.1 + y0 + y2 + 2y3 + G1(0) + 2y3G1(y3) 1.62/0.82 g = \y0.y0 1.62/0.82 h = \y0.y0 1.62/0.82 map = \G0y1.2 + 3y1 + 2y1G0(y1) + 2G0(y1) 1.62/0.82 nil = 0 1.62/0.82 true = 3 1.62/0.82 1.62/0.82 Using this interpretation, the requirements translate to: 1.62/0.82 1.62/0.82 [[f(g(_x0))]] = x0 >= x0 = [[g(f(f(_x0)))]] 1.62/0.82 [[f(h(_x0))]] = x0 >= x0 = [[h(g(_x0))]] 1.62/0.82 [[map(_F0, nil)]] = 2 + 2F0(0) > 0 = [[nil]] 1.62/0.82 [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 4F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 1.62/0.82 [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 1.62/0.82 [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 1 + 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 1.62/0.82 [[filter2(true, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 1.62/0.82 [[filter2(false, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 1.62/0.82 1.62/0.82 We can thus remove the following rules: 1.62/0.82 1.62/0.82 map(F, nil) => nil 1.62/0.82 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.62/0.82 filter(F, nil) => nil 1.62/0.82 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 1.62/0.82 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 1.62/0.82 filter2(false, F, X, Y) => filter(F, Y) 1.62/0.82 1.62/0.82 We observe that the rules contain a first-order subset: 1.62/0.82 1.62/0.82 f(g(X)) => g(f(f(X))) 1.62/0.82 f(h(X)) => h(g(X)) 1.62/0.82 1.62/0.82 Moreover, the system is orthogonal. 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 terminating when seen as a many-sorted first-order TRS. 1.62/0.82 1.62/0.82 According to the external first-order termination prover, this system is indeed terminating: 1.62/0.82 1.62/0.82 || proof of resources/system.trs 1.62/0.82 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.62/0.82 || 1.62/0.82 || 1.62/0.82 || Termination w.r.t. Q of the given QTRS could be proven: 1.62/0.82 || 1.62/0.82 || (0) QTRS 1.62/0.82 || (1) QTRS Reverse [EQUIVALENT] 1.62/0.82 || (2) QTRS 1.62/0.82 || (3) RFCMatchBoundsTRSProof [EQUIVALENT] 1.62/0.82 || (4) YES 1.62/0.82 || 1.62/0.82 || 1.62/0.82 || ----------------------------------------
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