Termination Proof Script

Consider the TRS R consisting of the rewrite rules


1:		active(f(b,c,x))	→ mark(f(x,x,x))
2:		active(f(x,y,z))	→ f(x,y,active(z))
3:		active(d)	→ m(b)
4:		f(x,y,mark(z))	→ mark(f(x,y,z))
5:		active(d)	→ mark(c)
6:		proper(b)	→ ok(b)
7:		proper(c)	→ ok(c)
8:		proper(d)	→ ok(d)
9:		proper(f(x,y,z))	→ f(proper(x),proper(y),proper(z))
10:		f(ok(x),ok(y),ok(z))	→ ok(f(x,y,z))
11:		top(mark(x))	→ top(proper(x))
12:		top(ok(x))	→ top(active(x))

There are 13 dependency pairs:


13:		ACTIVE(f(b,c,x))	→ F(x,x,x)
14:		ACTIVE(f(x,y,z))	→ F(x,y,active(z))
15:		ACTIVE(f(x,y,z))	→ ACTIVE(z)
16:		F(x,y,mark(z))	→ F(x,y,z)
17:		PROPER(f(x,y,z))	→ F(proper(x),proper(y),proper(z))
18:		PROPER(f(x,y,z))	→ PROPER(x)
19:		PROPER(f(x,y,z))	→ PROPER(y)
20:		PROPER(f(x,y,z))	→ PROPER(z)
21:		F(ok(x),ok(y),ok(z))	→ F(x,y,z)
22:		TOP(mark(x))	→ TOP(proper(x))
23:		TOP(mark(x))	→ PROPER(x)
24:		TOP(ok(x))	→ TOP(active(x))
25:		TOP(ok(x))	→ ACTIVE(x)

The approximated dependency graph contains 4 SCCs: {16,21}, {15}, {18-20} and {22,24}.

Consider the SCC {16,21}. There are no usable rules. By taking the AF π with π(F) = 1 together with the lexicographic path order with empty precedence, rule 16 is weakly decreasing and rule 21 is strictly decreasing. There is one new SCC.
- Consider the SCC {16}. By taking the AF π with π(F) = 3 together with the lexicographic path order with empty precedence, rule 16 is strictly decreasing.

Consider the SCC {15}. There are no usable rules. By taking the AF π with π(ACTIVE) = 1 and π(f) = [3] together with the lexicographic path order with empty precedence, rule 15 is strictly decreasing.

Consider the SCC {18-20}. There are no usable rules. By taking the AF π with π(PROPER) = 1 together with the lexicographic path order with empty precedence, the rules in {18-20} are strictly decreasing.

Consider the SCC {22,24}. The usable rules are {1-10}. The constraints could not be solved.

Tyrolean Termination Tool (0.97 seconds) --- May 3, 2006