Certification Problem

Input (TPDB TRS_Standard/Secret_07_TRS/secret1)

The rewrite relation of the following TRS is considered.

D(t)	→	s(h)	(1)
D(constant)	→	h	(2)
D(b(x,y))	→	b(D(x),D(y))	(3)
D(c(x,y))	→	b(c(y,D(x)),c(x,D(y)))	(4)
D(m(x,y))	→	m(D(x),D(y))	(5)
D(opp(x))	→	opp(D(x))	(6)
D(div(x,y))	→	m(div(D(x),y),div(c(x,D(y)),pow(y,2)))	(7)
D(ln(x))	→	div(D(x),x)	(8)
D(pow(x,y))	→	b(c(c(y,pow(x,m(y,1))),D(x)),c(c(pow(x,y),ln(x)),D(y)))	(9)
b(h,x)	→	x	(10)
b(x,h)	→	x	(11)
b(s(x),s(y))	→	s(s(b(x,y)))	(12)
b(b(x,y),z)	→	b(x,b(y,z))	(13)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

D^#(b(x,y))	→	D^#(y)	(14)
D^#(b(x,y))	→	D^#(x)	(15)
D^#(b(x,y))	→	b^#(D(x),D(y))	(16)
D^#(c(x,y))	→	D^#(y)	(17)
D^#(c(x,y))	→	D^#(x)	(18)
D^#(c(x,y))	→	b^#(c(y,D(x)),c(x,D(y)))	(19)
D^#(m(x,y))	→	D^#(y)	(20)
D^#(m(x,y))	→	D^#(x)	(21)
D^#(opp(x))	→	D^#(x)	(22)
D^#(div(x,y))	→	D^#(y)	(23)
D^#(div(x,y))	→	D^#(x)	(24)
D^#(ln(x))	→	D^#(x)	(25)
D^#(pow(x,y))	→	D^#(y)	(26)
D^#(pow(x,y))	→	D^#(x)	(27)
D^#(pow(x,y))	→	b^#(c(c(y,pow(x,m(y,1))),D(x)),c(c(pow(x,y),ln(x)),D(y)))	(28)
b^#(s(x),s(y))	→	b^#(x,y)	(29)
b^#(b(x,y),z)	→	b^#(y,z)	(30)
b^#(b(x,y),z)	→	b^#(x,b(y,z))	(31)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

D^#(ln(x))	→	D^#(x)	(25)
D^#(pow(x,y))	→	D^#(x)	(27)
D^#(pow(x,y))	→	D^#(y)	(26)
D^#(div(x,y))	→	D^#(x)	(24)
D^#(div(x,y))	→	D^#(y)	(23)
D^#(opp(x))	→	D^#(x)	(22)
D^#(m(x,y))	→	D^#(x)	(21)
D^#(m(x,y))	→	D^#(y)	(20)
D^#(c(x,y))	→	D^#(x)	(18)
D^#(c(x,y))	→	D^#(y)	(17)
D^#(b(x,y))	→	D^#(x)	(15)
D^#(b(x,y))	→	D^#(y)	(14)

1.1.1 Subterm Criterion Processor

We use the projection

π(D^#)

=

1

and remove the pairs:

D^#(ln(x))	→	D^#(x)	(25)
D^#(pow(x,y))	→	D^#(x)	(27)
D^#(pow(x,y))	→	D^#(y)	(26)
D^#(div(x,y))	→	D^#(x)	(24)
D^#(div(x,y))	→	D^#(y)	(23)
D^#(opp(x))	→	D^#(x)	(22)
D^#(m(x,y))	→	D^#(x)	(21)
D^#(m(x,y))	→	D^#(y)	(20)
D^#(c(x,y))	→	D^#(x)	(18)
D^#(c(x,y))	→	D^#(y)	(17)
D^#(b(x,y))	→	D^#(x)	(15)
D^#(b(x,y))	→	D^#(y)	(14)

1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

b^#(b(x,y),z) → b^#(y,z) (30)

b^#(b(x,y),z) → b^#(x,b(y,z)) (31)

b^#(s(x),s(y)) → b^#(x,y) (29)

1.1.2 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

b^#(b(x,y),z) → b^#(y,z) (30)

2 ≥ 2

1 > 1

b^#(b(x,y),z) → b^#(x,b(y,z)) (31)

1 > 1

b^#(s(x),s(y)) → b^#(x,y) (29)

2 > 2

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.