/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO

We split firstr-order part and higher-order part, and do modular checking by a general modularity.

******** FO SN check ********
Check non-SN using NTI (Non-Termination Inference by Payet)

******** Computation rules ********
 (1)  f(s(X)) => f(g(X,X))
 (2)  g(0,1) => s(0)
 (3)  0 => 1

forward+backward process killed -- 7 rule(s) generated

NO

let R be the TRS under consideration

f(s(_1)) -> f(g(_1,_1)) is in elim_R(R)
let l0 be the left-hand side of this rule
p0 = 0 is a position in l0
we have l0|p0 = s(_1)
g(0,1) -> s(0) is in R
let r'0 be the right-hand side of this rule
theta0 = {_1/0} is a mgu of l0|p0 and r'0

==> f(g(0,1)) -> f(g(0,0)) is in EU_R^1
let l1 be the left-hand side of this rule
p1 = 0.1 is a position in l1
we have l1|p1 = 1
0 -> 1 is in R
let r'1 be the right-hand side of this rule
theta1 = {} is a mgu of l1|p1 and r'1

==> f(g(0,0)) -> f(g(0,0)) is in EU_R^2
let l be the left-hand side and r be the right-hand side of this rule
let p = epsilon
let theta = {}
let theta' = {}
we have r|p = f(g(0,0)) and
theta'(theta(l)) = theta(r|p)
so, theta(l) = f(g(0,0)) is non-terminating w.r.t. R

Termination disproved by the backward process
proof stopped at iteration i=2, depth k=2
4 rule(s) generated
>>NO


******** Signature ********
 0 : b
 1 : b
 cons : (e,f) -> f
 f : c -> a
 false : d
 filter : ((e -> d),f) -> f
 filter2 : (d,(e -> d),e,f) -> f
 g : (b,b) -> c
 map : ((e -> e),f) -> f
 nil : f
 s : b -> c
 true : d
******** Computation Rules ********
 (1)  f(s(X)) => f(g(X,X))
 (2)  g(0,1) => s(0)
 (3)  0 => 1
 (4)  map(Z,nil) => nil
 (5)  map(G,cons(V,W)) => cons(G[V],map(G,W))
 (6)  filter(J,nil) => nil
 (7)  filter(F1,cons(Y1,U1)) => filter2(F1[Y1],F1,Y1,U1)
 (8)  filter2(true,H1,W1,P1) => cons(W1,filter(H1,P1))
 (9)  filter2(false,F2,Y2,U2) => filter(F2,U2)

NO