Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96224)

The rewrite relation of the following TRS is considered.

0(1(2(2(3(0(4(4(x1))))))))	→	0(4(5(0(3(4(0(4(x1))))))))	(1)
0(2(2(2(0(5(2(5(4(x1)))))))))	→	0(0(0(3(4(3(0(1(0(x1)))))))))	(2)
2(1(1(3(4(3(1(1(5(x1)))))))))	→	1(0(2(3(1(0(5(1(5(x1)))))))))	(3)
5(4(0(4(3(3(1(2(5(3(0(x1)))))))))))	→	5(5(2(5(3(1(5(0(3(5(2(x1)))))))))))	(4)
2(4(5(0(1(1(3(3(5(3(0(0(x1))))))))))))	→	4(4(2(2(1(0(4(0(1(3(2(0(x1))))))))))))	(5)
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x1))))))))))))))	→	0(2(4(5(0(0(0(5(1(0(5(4(4(x1)))))))))))))	(6)
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x1)))))))))))))))	→	0(0(3(5(3(0(4(3(1(3(0(2(5(5(x1))))))))))))))	(7)
3(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1)))))))))))))))))	→	3(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x1)))))))))))))))))	(8)
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x1)))))))))))))))))	→	5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x1)))))))))))))))))	(9)
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x1))))))))))))))))))	→	0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x1))))))))))))))))))	(10)
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x1))))))))))))))))))))	→	1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x1))))))))))))))))))))	(11)
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x1))))))))))))))))))))	→	5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x1)))))))))))))))))))	(12)
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x1))))))))))))))))))))	→	3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x1)))))))))))))))))))	(13)
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x1))))))))))))))))))))	→	3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x1))))))))))))))))))))	(14)
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x1))))))))))))))))))))	→	4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x1)))))))))))))))))))	(15)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals

[0(x₁)]	=	1 · x₁ + 104
[1(x₁)]	=	1 · x₁ + 131
[2(x₁)]	=	1 · x₁ + 126
[3(x₁)]	=	1 · x₁ + 164
[4(x₁)]	=	1 · x₁ + 138
[5(x₁)]	=	1 · x₁ + 139

all of the following rules can be deleted.

0(1(2(2(3(0(4(4(x1))))))))	→	0(4(5(0(3(4(0(4(x1))))))))	(1)
0(2(2(2(0(5(2(5(4(x1)))))))))	→	0(0(0(3(4(3(0(1(0(x1)))))))))	(2)
2(1(1(3(4(3(1(1(5(x1)))))))))	→	1(0(2(3(1(0(5(1(5(x1)))))))))	(3)
5(4(0(4(3(3(1(2(5(3(0(x1)))))))))))	→	5(5(2(5(3(1(5(0(3(5(2(x1)))))))))))	(4)
2(4(5(0(1(1(3(3(5(3(0(0(x1))))))))))))	→	4(4(2(2(1(0(4(0(1(3(2(0(x1))))))))))))	(5)
4(4(3(0(3(1(5(3(5(1(3(1(5(3(x1))))))))))))))	→	0(2(4(5(0(0(0(5(1(0(5(4(4(x1)))))))))))))	(6)
2(2(0(0(2(1(0(5(3(2(2(1(4(0(5(x1)))))))))))))))	→	0(0(3(5(3(0(4(3(1(3(0(2(5(5(x1))))))))))))))	(7)
5(1(1(4(1(5(3(0(4(3(2(5(4(1(3(3(5(x1)))))))))))))))))	→	5(0(1(0(4(0(2(4(5(1(5(4(1(5(3(3(5(x1)))))))))))))))))	(9)
0(1(4(4(3(2(0(4(1(4(3(4(4(1(5(3(4(4(x1))))))))))))))))))	→	0(2(4(2(3(1(0(1(1(1(3(0(2(4(4(1(1(2(x1))))))))))))))))))	(10)
1(1(4(1(0(1(0(3(3(4(4(1(5(4(0(4(4(5(5(3(x1))))))))))))))))))))	→	1(3(0(3(2(2(4(4(2(0(3(3(4(0(3(0(4(3(4(0(x1))))))))))))))))))))	(11)
2(0(3(3(3(4(1(1(0(4(4(0(3(3(3(0(0(1(5(3(x1))))))))))))))))))))	→	5(3(3(1(4(0(4(5(4(4(4(2(4(3(1(1(1(5(4(x1)))))))))))))))))))	(12)
2(3(3(2(1(5(0(5(0(1(3(3(2(5(1(5(0(3(0(5(x1))))))))))))))))))))	→	3(4(0(2(5(5(2(4(2(4(3(1(1(4(4(5(5(3(5(x1)))))))))))))))))))	(13)
3(3(2(2(3(3(4(0(0(0(2(5(0(5(3(0(0(1(1(4(x1))))))))))))))))))))	→	3(1(1(0(3(3(5(4(2(2(1(0(1(1(0(5(0(3(4(0(x1))))))))))))))))))))	(14)
4(2(4(1(0(5(0(4(1(0(3(0(2(5(4(3(5(3(5(3(x1))))))))))))))))))))	→	4(2(5(4(4(2(4(5(0(0(3(1(5(0(2(0(2(2(1(x1)))))))))))))))))))	(15)

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

3^#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1)))))))))))))))))	→	3^#(4(0(1(0(5(5(3(1(4(0(3(5(3(2(2(5(x1)))))))))))))))))	(16)
3^#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1)))))))))))))))))	→	3^#(1(4(0(3(5(3(2(2(5(x1))))))))))	(17)
3^#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1)))))))))))))))))	→	3^#(5(3(2(2(5(x1))))))	(18)
3^#(0(0(4(2(5(5(1(3(0(2(3(3(5(1(4(5(x1)))))))))))))))))	→	3^#(2(2(5(x1))))	(19)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.