YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

f(g(x,a,b)) → x
g(f(h(c,d)),x,y) → h(k1(x),k2(y))
k1(a) → c
k2(b) → d
f(h(k1(a),k2(b))) → f(h(c,d))
f(h(c,k2(b))) → f(h(c,d))
f(h(k1(a),d)) → f(h(c,d))

Proof

1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

f(h(k1(a),k2(b))) → f(h(c,d))
k2(b) → d
k1(a) → c
g(f(h(c,d)),x,y) → h(k1(x),k2(y))
f(g(x,a,b)) → x

All redundant rules that were added or removed can be simulated in 1 steps .

1.1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

f(h(k1(a),k2(b))) → f(h(c,d))
k2(b) → d
k1(a) → c
g(f(h(c,d)),x,y) → h(k1(x),k2(y))
f(g(x,a,b)) → x

All redundant rules that were added or removed can be simulated in 2 steps .

1.1.1 Critical Pair Closing System

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

k1(a) → c
k2(b) → d
f(h(k1(a),k2(b))) → f(h(c,d))

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[k2(x1)] = 4 · x1 + 0
[b] = 0
[f(x1)] = 1 · x1 + 0
[d] = 0
[a] = 1
[k1(x1)] = 1 · x1 + 1
[h(x1, x2)] = 2 · x1 + 1 · x2 + 3
the rule
k2(b) → d
remains.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[k2(x1)] = 1 · x1 + 1
[b] = 4
[d] = 0
all rules could be removed.

1.1.1.1.1.1 R is empty

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

Tool configuration

csi