(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(++(x1, x2)) = 2 + x1 + x2
POL(.(x1, x2)) = x1 + x2
POL(make(x1)) = 1 + 2·x1
POL(nil) = 0
POL(rev(x1)) = 1 + 2·x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
make(x) → .(x, nil)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(++(x1, x2)) = 1 + x1 + 2·x2
POL(.(x1, x2)) = x1 + x2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
++(x, ++(y, z)) → ++(++(x, y), z)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
++(.(x, y), z) → .(x, ++(y, z))
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Knuth-Bendix order [KBO] with precedence:
++2 > .2
and weight map:
._2=0
++_2=0
The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
++(.(x, y), z) → .(x, ++(y, z))
(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