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:
+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)
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(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)
Used ordering:
Polynomial interpretation [25]:
POL(+(x1, x2)) = x1 + x2
POL(f(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(+(x, y), z) → +(x, +(y, z))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
+(+(x, y), z) → +(x, +(y, z))
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, y), z) → +(x, +(y, z))
Used ordering:
Polynomial interpretation [25]:
POL(+(x1, x2)) = 2 + 2·x1 + x2
↳ 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.