Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-484)

The rewrite relation of the following TRS is considered.

a(a(x1))	→	a(b(c(b(x1))))	(1)
b(c(x1))	→	x1	(2)
c(b(x1))	→	a(c(x1))	(3)

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(x1))	→	b(c(b(a(x1))))	(4)
c(b(x1))	→	x1	(5)
b(c(x1))	→	c(a(x1))	(6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(x1))	→	b^#(a(x1))	(7)
a^#(a(x1))	→	c^#(b(a(x1)))	(8)
a^#(a(x1))	→	b^#(c(b(a(x1))))	(9)
b^#(c(x1))	→	a^#(x1)	(10)
b^#(c(x1))	→	c^#(a(x1))	(11)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(c(x1))	→	a^#(x1)	(10)
a^#(a(x1))	→	b^#(a(x1))	(7)
a^#(a(x1))	→	b^#(c(b(a(x1))))	(9)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers

[b^#(x₁)]

1	2
-∞	-∞

· x₁ +

0	-∞
-∞	-∞

[b(x₁)]

-∞	0
0	1

· x₁ +

0	-∞
0	-∞

[a^#(x₁)]

1	0
-∞	-∞

· x₁ +

0	-∞
-∞	-∞

[a(x₁)]

1	0
0	-∞

· x₁ +

1	-∞
0	-∞

[c(x₁)]

-∞	0
0	-∞

· x₁ +

0	-∞
0	-∞

together with the usable rules

a(a(x1))	→	b(c(b(a(x1))))	(4)
c(b(x1))	→	x1	(5)
b(c(x1))	→	c(a(x1))	(6)

(w.r.t. the implicit argument filter of the reduction pair), the pair

b^#(c(x1))

→

a^#(x1)

(10)

could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.