(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(cons(x1, x2)) = 1 + x1 + x2
POL(empty) = 0
POL(f(x1, x2)) = 1 + 2·x1 + 2·x2
POL(g(x1, x2)) = 2·x1 + x2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(a, empty) → g(a, empty)
g(cons(x, k), d) → g(k, cons(x, d))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(cons(x1, x2)) = x1 + x2
POL(empty) = 0
POL(f(x1, x2)) = 2·x1 + 2·x2
POL(g(x1, x2)) = 2 + x1 + x2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
g(empty, d) → d
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, cons(x, k)) → f(cons(x, a), k)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(cons(x1, x2)) = 1 + x1 + x2
POL(f(x1, x2)) = x1 + 2·x2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(a, cons(x, k)) → f(cons(x, a), k)
(6) Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.
(7) RisEmptyProof (EQUIVALENT transformation)
The TRS R is empty. Hence, termination is trivially proven.
(8) YES