YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))
c c
f(f(h(h(f(a),a),c))) f(h(f(c),b))
h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) c

Proof

1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))
f(f(h(h(f(a),a),c))) f(h(f(c),b))
h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) c

1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[a] =
0 0
1 0
[f(x1)] =
1 1
0 0
· x1 +
0 0
0 0
[h(x1, x2)] =
2 0
0 0
· x1 +
2 0
0 0
· x2 +
0 0
0 0
[c] =
0 0
0 0
[b] =
0 0
0 0
the rules
h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))
h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) c
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[a] =
0 0
1 0
[f(x1)] =
2 1
0 0
· x1 +
0 0
0 0
[h(x1, x2)] =
2 0
1 0
· x1 +
2 0
0 1
· x2 +
0 0
0 0
[c] =
0 0
0 0
[b] =
0 0
0 0
the rule
h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))
remains.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[a] =
0 0
0 0
[f(x1)] =
1 2
0 1
· x1 +
0 0
0 0
[h(x1, x2)] =
1 0
0 0
· x1 +
2 3
0 0
· x2 +
0 0
0 0
[c] =
0 0
1 0
[b] =
0 0
0 0
all rules could be removed.

1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.

Tool configuration

csi