Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/matchbox1)

The rewrite relation of the following TRS is considered.

r(e(x1)) → w(r(x1)) (1)
i(t(x1)) → e(r(x1)) (2)
e(w(x1)) → r(i(x1)) (3)
t(e(x1)) → r(e(x1)) (4)
w(r(x1)) → i(t(x1)) (5)
e(r(x1)) → e(w(x1)) (6)
r(i(t(e(r(x1))))) → e(w(r(i(t(e(x1)))))) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
i#(t(x1)) → r#(x1) (8)
r#(i(t(e(r(x1))))) → w#(r(i(t(e(x1))))) (9)
e#(w(x1)) → r#(i(x1)) (10)
r#(i(t(e(r(x1))))) → t#(e(x1)) (11)
w#(r(x1)) → i#(t(x1)) (12)
w#(r(x1)) → t#(x1) (13)
e#(w(x1)) → i#(x1) (14)
r#(i(t(e(r(x1))))) → r#(i(t(e(x1)))) (15)
r#(i(t(e(r(x1))))) → e#(x1) (16)
r#(i(t(e(r(x1))))) → e#(w(r(i(t(e(x1)))))) (17)
t#(e(x1)) → r#(e(x1)) (18)
r#(e(x1)) → r#(x1) (19)
e#(r(x1)) → e#(w(x1)) (20)
i#(t(x1)) → e#(r(x1)) (21)
r#(i(t(e(r(x1))))) → i#(t(e(x1))) (22)
r#(e(x1)) → w#(r(x1)) (23)
e#(r(x1)) → w#(x1) (24)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.