Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
a__b → b
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
a__b → b
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(b) → a__b
mark(a) → a
a__f(X1, X2, X3) → f(X1, X2, X3)
a__b → b
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
mark(b) → a__b
mark(a) → a
a__b → b
Used ordering:
Polynomial interpretation [25]:
POL(a) = 1
POL(a__b) = 1
POL(a__f(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3
POL(b) = 0
POL(f(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + x3
POL(mark(x1)) = 2 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
a__f(X1, X2, X3) → f(X1, X2, X3)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
a__f(X1, X2, X3) → f(X1, X2, X3)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
a__f(X1, X2, X3) → f(X1, X2, X3)
Used ordering:
Polynomial interpretation [25]:
POL(a) = 0
POL(a__b) = 0
POL(a__f(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3
POL(b) = 0
POL(f(x1, x2, x3)) = 1 + x1 + 2·x2 + 2·x3
POL(mark(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
Used ordering:
Polynomial interpretation [25]:
POL(a) = 0
POL(a__b) = 0
POL(a__f(x1, x2, x3)) = x1 + 2·x2 + x3
POL(b) = 0
POL(f(x1, x2, x3)) = 1 + 2·x1 + 2·x2 + 2·x3
POL(mark(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
a__b → a
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
a__b → a
Used ordering:
Polynomial interpretation [25]:
POL(a) = 1
POL(a__b) = 2
POL(a__f(x1, x2, x3)) = 2·x1 + x2 + x3
POL(b) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, X, X) → a__f(X, a__b, b)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
a__f(a, X, X) → a__f(X, a__b, b)
Used ordering:
Polynomial interpretation [25]:
POL(a) = 2
POL(a__b) = 1
POL(a__f(x1, x2, x3)) = 2·x1 + 2·x2 + x3
POL(b) = 1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RisEmptyProof
Q restricted rewrite system:
R is empty.
Q is empty.
The TRS R is empty. Hence, termination is trivially proven.