Certification Problem

Input (TPDB SRS_Standard/Zantema_06/beans2)

The rewrite relation of the following TRS is considered.

b(a(a(x1)))	→	a(b(c(x1)))	(1)
c(a(x1))	→	a(c(x1))	(2)
c(b(x1))	→	b(a(x1))	(3)
L(a(a(x1)))	→	L(a(b(c(x1))))	(4)
c(R(x1))	→	b(a(R(x1)))	(5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

a(a(b(x1)))	→	c(b(a(x1)))	(6)
a(c(x1))	→	c(a(x1))	(7)
b(c(x1))	→	a(b(x1))	(8)
a(a(L(x1)))	→	c(b(a(L(x1))))	(9)
R(c(x1))	→	R(a(b(x1)))	(10)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(b(x1)))	→	a^#(x1)	(11)
a^#(a(b(x1)))	→	b^#(a(x1))	(12)
a^#(c(x1))	→	a^#(x1)	(13)
b^#(c(x1))	→	b^#(x1)	(14)
b^#(c(x1))	→	a^#(b(x1))	(15)
a^#(a(L(x1)))	→	b^#(a(L(x1)))	(16)
R^#(c(x1))	→	b^#(x1)	(17)
R^#(c(x1))	→	a^#(b(x1))	(18)
R^#(c(x1))	→	R^#(a(b(x1)))	(19)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
R^#(c(x1)) → R^#(a(b(x1))) (19)

1.1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers

[b(x₁)] = -1 · x₁ + 0

[L(x₁)] = -∞ · x₁ + 2

[a(x₁)] = 2 · x₁ + 3

[R^#(x₁)] = 0 · x₁ + 0

[c(x₁)] = 2 · x₁ + 4

together with the usable rules

b(c(x1)) → a(b(x1)) (8)

a(a(b(x1))) → c(b(a(x1))) (6)

a(c(x1)) → c(a(x1)) (7)

a(a(L(x1))) → c(b(a(L(x1)))) (9)

(w.r.t. the implicit argument filter of the reduction pair), the pair
R^#(c(x1)) → R^#(a(b(x1))) (19)
could be deleted.
1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

a^#(c(x1))	→	a^#(x1)	(13)
a^#(a(b(x1)))	→	b^#(a(x1))	(12)
b^#(c(x1))	→	a^#(b(x1))	(15)
a^#(a(b(x1)))	→	a^#(x1)	(11)
b^#(c(x1))	→	b^#(x1)	(14)

1.1.1.2 Subterm Criterion Processor

We use the projection to multisets

π(b^#)	=	{ 1 }
π(a^#)	=	{ 1, 1 }
π(c)	=	{ 1, 1, 1 }
π(b)	=	{ 1 }
π(a)	=	{ 1, 1, 1 }

to remove the pairs:

a^#(c(x1))	→	a^#(x1)	(13)
a^#(a(b(x1)))	→	b^#(a(x1))	(12)
b^#(c(x1))	→	a^#(b(x1))	(15)
a^#(a(b(x1)))	→	a^#(x1)	(11)
b^#(c(x1))	→	b^#(x1)	(14)

1.1.1.2.1 P is empty

There are no pairs anymore.