YES
by csi
The rewrite relation of the following TRS is considered.
a(b(a(x))) | → | b(a(b(x))) |
b(a(b(x))) | → | a(b(a(x))) |
p | → | a(p) |
p | → | b(p) |
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
p | → | b(p) |
p | → | a(p) |
b(a(b(x))) | → | a(b(a(x))) |
a(b(a(x))) | → | b(a(b(x))) |
p | → | b(b(p)) |
p | → | b(a(p)) |
p | → | a(b(p)) |
p | → | a(a(p)) |
b(a(b(x))) | → | b(a(b(x))) |
a(b(a(x))) | → | a(b(a(x))) |
All redundant rules that were added or removed can be simulated in 2 steps .
The duplicating rules (R) terminate relative to the other rules (S).
There are no rules in the TRS R. Hence, R/S is relative terminating.
csi