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:
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
++(x, nil) → x
++(x, g(y, z)) → g(++(x, y), z)
null(nil) → true
null(g(x, y)) → false
mem(nil, y) → false
mem(g(x, y), z) → or(=(y, z), mem(x, z))
mem(x, max(x)) → not(null(x))
max(g(g(nil, x), y)) → max'(x, y)
max(g(g(g(x, y), z), u)) → max'(max(g(g(x, y), z)), u)
Q is empty.
↳ QTRS
↳ DirectTerminationProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
++(x, nil) → x
++(x, g(y, z)) → g(++(x, y), z)
null(nil) → true
null(g(x, y)) → false
mem(nil, y) → false
mem(g(x, y), z) → or(=(y, z), mem(x, z))
mem(x, max(x)) → not(null(x))
max(g(g(nil, x), y)) → max'(x, y)
max(g(g(g(x, y), z), u)) → max'(max(g(g(x, y), z)), u)
Q is empty.
We use [23] with the following order to prove termination.
Recursive Path Order [2].
Precedence:
f2 > g2
++2 > g2
mem2 > null1 > true
mem2 > null1 > false
mem2 > or2
mem2 > =2
mem2 > not1
max1 > max'2