Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/cime1)

The rewrite relation of the following TRS is considered.

sortSu(circ(sortSu(cons(te(a),sortSu(s))),sortSu(t))) sortSu(cons(te(msubst(te(a),sortSu(t))),sortSu(circ(sortSu(s),sortSu(t))))) (1)
sortSu(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(cons(te(a),sortSu(t))))) sortSu(cons(te(a),sortSu(circ(sortSu(s),sortSu(t))))) (2)
sortSu(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(cons(sop(lift),sortSu(t))))) sortSu(cons(sop(lift),sortSu(circ(sortSu(s),sortSu(t))))) (3)
sortSu(circ(sortSu(circ(sortSu(s),sortSu(t))),sortSu(u))) sortSu(circ(sortSu(s),sortSu(circ(sortSu(t),sortSu(u))))) (4)
sortSu(circ(sortSu(s),sortSu(id))) sortSu(s) (5)
sortSu(circ(sortSu(id),sortSu(s))) sortSu(s) (6)
sortSu(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(circ(sortSu(cons(sop(lift),sortSu(t))),sortSu(u))))) sortSu(circ(sortSu(cons(sop(lift),sortSu(circ(sortSu(s),sortSu(t))))),sortSu(u))) (7)
te(subst(te(a),sortSu(id))) te(a) (8)
te(msubst(te(a),sortSu(id))) te(a) (9)
te(msubst(te(msubst(te(a),sortSu(s))),sortSu(t))) te(msubst(te(a),sortSu(circ(sortSu(s),sortSu(t))))) (10)

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.
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(cons(te(a),sortSu(t))))) sortSu#(circ(sortSu(s),sortSu(t))) (11)
te#(msubst(te(msubst(te(a),sortSu(s))),sortSu(t))) te#(msubst(te(a),sortSu(circ(sortSu(s),sortSu(t))))) (12)
sortSu#(circ(sortSu(circ(sortSu(s),sortSu(t))),sortSu(u))) sortSu#(circ(sortSu(s),sortSu(circ(sortSu(t),sortSu(u))))) (13)
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(cons(sop(lift),sortSu(t))))) sortSu#(circ(sortSu(s),sortSu(t))) (14)
sortSu#(circ(sortSu(cons(te(a),sortSu(s))),sortSu(t))) te#(msubst(te(a),sortSu(t))) (15)
sortSu#(circ(sortSu(cons(te(a),sortSu(s))),sortSu(t))) sortSu#(circ(sortSu(s),sortSu(t))) (16)
sortSu#(circ(sortSu(circ(sortSu(s),sortSu(t))),sortSu(u))) sortSu#(circ(sortSu(t),sortSu(u))) (17)
te#(msubst(te(msubst(te(a),sortSu(s))),sortSu(t))) sortSu#(circ(sortSu(s),sortSu(t))) (18)
sortSu#(circ(sortSu(cons(te(a),sortSu(s))),sortSu(t))) sortSu#(cons(te(msubst(te(a),sortSu(t))),sortSu(circ(sortSu(s),sortSu(t))))) (19)
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(circ(sortSu(cons(sop(lift),sortSu(t))),sortSu(u))))) sortSu#(circ(sortSu(s),sortSu(t))) (20)
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(circ(sortSu(cons(sop(lift),sortSu(t))),sortSu(u))))) sortSu#(circ(sortSu(cons(sop(lift),sortSu(circ(sortSu(s),sortSu(t))))),sortSu(u))) (21)
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(cons(te(a),sortSu(t))))) sortSu#(cons(te(a),sortSu(circ(sortSu(s),sortSu(t))))) (22)
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(cons(sop(lift),sortSu(t))))) sortSu#(cons(sop(lift),sortSu(circ(sortSu(s),sortSu(t))))) (23)
sortSu#(circ(sortSu(cons(sop(lift),sortSu(s))),sortSu(circ(sortSu(cons(sop(lift),sortSu(t))),sortSu(u))))) sortSu#(cons(sop(lift),sortSu(circ(sortSu(s),sortSu(t))))) (24)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.