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:
g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
g(C) → A
g(C) → B
Used ordering:
Polynomial interpretation [25]:
POL(A) = 0
POL(B) = 0
POL(C) = 1
POL(cons(x1, x2)) = x1 + x2
POL(f(x1, x2)) = x1 + x2
POL(f'(x1, x2)) = x1 + x2
POL(f''(x1)) = x1
POL(foldf(x1, x2)) = x1 + x2
POL(g(x1)) = x1
POL(nil) = 0
POL(triple(x1, x2, x3)) = 2·x1 + x2 + x3
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → A
g(B) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → A
g(B) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
g(B) → A
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
Used ordering:
Polynomial interpretation [25]:
POL(A) = 0
POL(B) = 1
POL(C) = 0
POL(cons(x1, x2)) = 2·x1 + x2
POL(f(x1, x2)) = x1 + 2·x2
POL(f'(x1, x2)) = x1 + 2·x2
POL(f''(x1)) = x1
POL(foldf(x1, x2)) = x1 + x2
POL(g(x1)) = x1
POL(nil) = 0
POL(triple(x1, x2, x3)) = 2·x1 + 2·x2 + x3
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
Used ordering:
Polynomial interpretation [25]:
POL(A) = 2
POL(B) = 0
POL(C) = 0
POL(cons(x1, x2)) = x1 + x2
POL(f(x1, x2)) = x1 + 2·x2
POL(f'(x1, x2)) = x1 + 2·x2
POL(f''(x1)) = 1 + x1
POL(foldf(x1, x2)) = x1 + 2·x2
POL(g(x1)) = x1
POL(nil) = 0
POL(triple(x1, x2, x3)) = x1 + 2·x2 + 2·x3
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(A) → A
g(B) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
g(A) → A
g(B) → B
foldf(x, nil) → x
Used ordering:
Polynomial interpretation [25]:
POL(A) = 2
POL(B) = 2
POL(C) = 0
POL(cons(x1, x2)) = x1 + x2
POL(f(x1, x2)) = x1 + 2·x2
POL(f'(x1, x2)) = x1 + x2
POL(foldf(x1, x2)) = 2·x1 + 2·x2
POL(g(x1)) = 2·x1
POL(nil) = 2
POL(triple(x1, x2, x3)) = 2·x1 + 2·x2 + x3
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(C) → C
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(C) → C
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
Used ordering:
Polynomial interpretation [25]:
POL(C) = 2
POL(cons(x1, x2)) = x1 + x2
POL(f(x1, x2)) = x1 + 2·x2
POL(f'(x1, x2)) = x1 + 2·x2
POL(foldf(x1, x2)) = x1 + 2·x2
POL(g(x1)) = x1
POL(triple(x1, x2, x3)) = 2·x1 + 2·x2 + x3
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(C) → C
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(C) → C
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
g(C) → C
Used ordering:
Polynomial interpretation [25]:
POL(C) = 2
POL(cons(x1, x2)) = x1 + 2·x2
POL(f(x1, x2)) = x1 + 2·x2
POL(f'(x1, x2)) = x1 + x2
POL(foldf(x1, x2)) = x1 + 2·x2
POL(g(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
Used ordering:
Polynomial interpretation [25]:
POL(cons(x1, x2)) = 1 + 2·x1 + 2·x2
POL(f(x1, x2)) = 1 + x1 + 2·x2
POL(f'(x1, x2)) = x1 + x2
POL(foldf(x1, x2)) = 2·x1 + 2·x2
POL(g(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ 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.