(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:
f3 > [0, s1]
Status:
f3: [1,2,3]
0: multiset
s1: multiset
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))
(2) Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.
(3) RisEmptyProof (EQUIVALENT transformation)
The TRS R is empty. Hence, termination is trivially proven.
(4) YES