YES
Termination w.r.t. Q proof of ../tpdb/qualif/bin.trs
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:
0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
0(#) → #
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
Used ordering:
Polynomial interpretation [25]:
POL(#) = 0
POL(+(x1, x2)) = x1 + 2·x2
POL(0(x1)) = 1 + x1
POL(1(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(#, x) → x
+(x, #) → x
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
+(#, x) → x
+(x, #) → x
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
Used ordering:
Polynomial interpretation [25]:
POL(#) = 0
POL(+(x1, x2)) = 2·x1 + 2·x2
POL(0(x1)) = 1 + x1
POL(1(x1)) = 1 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(#, x) → x
+(x, #) → x
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
+(#, x) → x
+(x, #) → 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:
+(#, x) → x
+(x, #) → x
Used ordering:
Polynomial interpretation [25]:
POL(#) = 2
POL(+(x1, x2)) = 2 + 2·x1 + 2·x2
↳ 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.