YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 Redundant Rules Transformation

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

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

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

1.1 Critical Pair Closing System

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

h(x,y) → c
g(b) → c
f(x) → c
f(x) → h(x,x)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[g(x1)] = 1 · x1 + 1
[b] = 3
[h(x1, x2)] = 4 · x1 + 2 · x2 + 0
[f(x1)] = 6 · x1 + 0
the rules
h(x,y) → c
f(x) → c
f(x) → h(x,x)
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[h(x1, x2)] = 2 · x1 + 2 · x2 + 0
[f(x1)] = 6 · x1 + 2
the rule
h(x,y) → c
remains.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[h(x1, x2)] = 4 · x1 + 4 · x2 + 2
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