Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212774)

The rewrite relation of the following TRS is considered.

0(0(1(2(x1))))	→	0(2(0(1(1(x1)))))	(1)
0(1(2(2(x1))))	→	0(2(2(1(0(x1)))))	(2)
0(1(2(2(x1))))	→	1(0(2(2(0(x1)))))	(3)
0(1(2(3(x1))))	→	0(2(0(1(3(x1)))))	(4)
0(1(2(3(x1))))	→	0(2(3(3(1(x1)))))	(5)
0(2(1(2(x1))))	→	0(2(2(0(1(x1)))))	(6)
0(2(1(2(x1))))	→	0(2(2(2(1(x1)))))	(7)
0(2(1(2(x1))))	→	0(2(2(2(1(1(x1))))))	(8)
0(3(2(2(x1))))	→	3(4(0(2(2(x1)))))	(9)
0(3(2(2(x1))))	→	0(2(2(2(2(3(x1))))))	(10)
0(4(1(2(x1))))	→	0(2(2(1(4(x1)))))	(11)
0(4(1(2(x1))))	→	4(0(2(0(1(x1)))))	(12)
0(5(0(1(x1))))	→	0(2(0(2(5(1(x1))))))	(13)
0(5(0(5(x1))))	→	0(2(0(5(5(x1)))))	(14)
0(5(2(1(x1))))	→	0(2(2(5(1(x1)))))	(15)
0(5(2(5(x1))))	→	0(2(2(5(5(x1)))))	(16)
0(5(4(2(x1))))	→	4(0(2(2(0(5(x1))))))	(17)
2(1(0(3(x1))))	→	4(0(2(2(3(1(x1))))))	(18)
2(1(0(4(x1))))	→	1(4(0(2(2(2(x1))))))	(19)
2(5(4(2(x1))))	→	4(0(2(2(5(x1)))))	(20)
0(0(1(0(4(x1)))))	→	0(0(2(0(1(4(x1))))))	(21)
0(0(5(4(2(x1)))))	→	0(4(0(0(2(5(x1))))))	(22)
0(1(0(1(2(x1)))))	→	0(2(0(1(4(1(x1))))))	(23)
0(1(2(0(3(x1)))))	→	0(2(0(4(1(3(x1))))))	(24)
0(1(2(2(2(x1)))))	→	0(2(2(2(1(2(x1))))))	(25)
0(1(2(3(2(x1)))))	→	1(3(4(0(2(2(x1))))))	(26)
0(1(3(2(3(x1)))))	→	0(0(2(3(3(1(x1))))))	(27)
0(1(3(4(2(x1)))))	→	0(2(3(4(1(1(x1))))))	(28)
0(2(1(0(1(x1)))))	→	0(0(2(0(1(1(x1))))))	(29)
0(2(1(2(2(x1)))))	→	0(2(0(2(2(1(x1))))))	(30)
0(2(3(0(5(x1)))))	→	0(2(0(0(5(3(x1))))))	(31)
0(3(0(1(3(x1)))))	→	0(0(4(3(1(3(x1))))))	(32)
0(3(0(4(1(x1)))))	→	0(0(1(4(4(3(x1))))))	(33)
0(3(2(0(4(x1)))))	→	4(0(0(2(3(4(x1))))))	(34)
0(4(5(2(3(x1)))))	→	0(2(2(3(4(5(x1))))))	(35)
0(5(0(0(3(x1)))))	→	0(2(0(3(0(5(x1))))))	(36)
0(5(0(1(2(x1)))))	→	0(0(2(0(1(5(x1))))))	(37)
0(5(1(4(2(x1)))))	→	0(2(0(1(4(5(x1))))))	(38)
0(5(2(5(1(x1)))))	→	0(2(0(5(5(1(x1))))))	(39)
2(1(0(0(4(x1)))))	→	1(4(4(0(0(2(x1))))))	(40)
2(5(0(0(3(x1)))))	→	0(2(0(0(5(3(x1))))))	(41)
2(5(3(0(1(x1)))))	→	5(0(2(2(3(1(x1))))))	(42)
5(0(1(2(2(x1)))))	→	5(1(0(2(0(2(x1))))))	(43)
5(2(0(1(2(x1)))))	→	1(5(4(0(2(2(x1))))))	(44)
5(2(1(0(1(x1)))))	→	0(2(3(1(5(1(x1))))))	(45)
5(2(3(0(1(x1)))))	→	1(5(0(2(2(3(x1))))))	(46)
5(3(0(4(1(x1)))))	→	4(5(0(2(3(1(x1))))))	(47)

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

2(1(0(0(x1))))	→	1(1(0(2(0(x1)))))	(48)
2(2(1(0(x1))))	→	0(1(2(2(0(x1)))))	(49)
2(2(1(0(x1))))	→	0(2(2(0(1(x1)))))	(50)
3(2(1(0(x1))))	→	3(1(0(2(0(x1)))))	(51)
3(2(1(0(x1))))	→	1(3(3(2(0(x1)))))	(52)
2(1(2(0(x1))))	→	1(0(2(2(0(x1)))))	(53)
2(1(2(0(x1))))	→	1(2(2(2(0(x1)))))	(54)
2(1(2(0(x1))))	→	1(1(2(2(2(0(x1))))))	(55)
2(2(3(0(x1))))	→	2(2(0(4(3(x1)))))	(56)
2(2(3(0(x1))))	→	3(2(2(2(2(0(x1))))))	(57)
2(1(4(0(x1))))	→	4(1(2(2(0(x1)))))	(58)
2(1(4(0(x1))))	→	1(0(2(0(4(x1)))))	(59)
1(0(5(0(x1))))	→	1(5(2(0(2(0(x1))))))	(60)
5(0(5(0(x1))))	→	5(5(0(2(0(x1)))))	(61)
1(2(5(0(x1))))	→	1(5(2(2(0(x1)))))	(62)
5(2(5(0(x1))))	→	5(5(2(2(0(x1)))))	(63)
2(4(5(0(x1))))	→	5(0(2(2(0(4(x1))))))	(64)
3(0(1(2(x1))))	→	1(3(2(2(0(4(x1))))))	(65)
4(0(1(2(x1))))	→	2(2(2(0(4(1(x1))))))	(66)
2(4(5(2(x1))))	→	5(2(2(0(4(x1)))))	(67)
4(0(1(0(0(x1)))))	→	4(1(0(2(0(0(x1))))))	(68)
2(4(5(0(0(x1)))))	→	5(2(0(0(4(0(x1))))))	(69)
2(1(0(1(0(x1)))))	→	1(4(1(0(2(0(x1))))))	(70)
3(0(2(1(0(x1)))))	→	3(1(4(0(2(0(x1))))))	(71)
2(2(2(1(0(x1)))))	→	2(1(2(2(2(0(x1))))))	(72)
2(3(2(1(0(x1)))))	→	2(2(0(4(3(1(x1))))))	(73)
3(2(3(1(0(x1)))))	→	1(3(3(2(0(0(x1))))))	(74)
2(4(3(1(0(x1)))))	→	1(1(4(3(2(0(x1))))))	(75)
1(0(1(2(0(x1)))))	→	1(1(0(2(0(0(x1))))))	(76)
2(2(1(2(0(x1)))))	→	1(2(2(0(2(0(x1))))))	(77)
5(0(3(2(0(x1)))))	→	3(5(0(0(2(0(x1))))))	(78)
3(1(0(3(0(x1)))))	→	3(1(3(4(0(0(x1))))))	(79)
1(4(0(3(0(x1)))))	→	3(4(4(1(0(0(x1))))))	(80)
4(0(2(3(0(x1)))))	→	4(3(2(0(0(4(x1))))))	(81)
3(2(5(4(0(x1)))))	→	5(4(3(2(2(0(x1))))))	(82)
3(0(0(5(0(x1)))))	→	5(0(3(0(2(0(x1))))))	(83)
2(1(0(5(0(x1)))))	→	5(1(0(2(0(0(x1))))))	(84)
2(4(1(5(0(x1)))))	→	5(4(1(0(2(0(x1))))))	(85)
1(5(2(5(0(x1)))))	→	1(5(5(0(2(0(x1))))))	(86)
4(0(0(1(2(x1)))))	→	2(0(0(4(4(1(x1))))))	(87)
3(0(0(5(2(x1)))))	→	3(5(0(0(2(0(x1))))))	(88)
1(0(3(5(2(x1)))))	→	1(3(2(2(0(5(x1))))))	(89)
2(2(1(0(5(x1)))))	→	2(0(2(0(1(5(x1))))))	(90)
2(1(0(2(5(x1)))))	→	2(2(0(4(5(1(x1))))))	(91)
1(0(1(2(5(x1)))))	→	1(5(1(3(2(0(x1))))))	(92)
1(0(3(2(5(x1)))))	→	3(2(2(0(5(1(x1))))))	(93)
1(4(0(3(5(x1)))))	→	1(3(2(0(5(4(x1))))))	(94)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 176 ruless (increase limit for explicit display).

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

3^#(1(0(3(0(x1)))))	→	3^#(4(0(0(x1))))	(206)
3^#(1(0(3(0(x1)))))	→	3^#(1(3(4(0(0(x1))))))	(208)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(3^#)	=	0	stat(3^#)	=	lex
prec(5)	=	0	stat(5)	=	lex
prec(4)	=	0	stat(4)	=	lex
prec(3)	=	0	stat(3)	=	lex
prec(0)	=	0	stat(0)	=	lex
prec(1)	=	1	stat(1)	=	lex
prec(2)	=	0	stat(2)	=	lex

π(3^#)	=	1
π(5)	=	[]
π(4)	=	[]
π(3)	=	1
π(0)	=	1
π(1)	=	[]
π(2)	=	1

together with the usable rules

4(0(0(1(2(x1)))))	→	2(0(0(4(4(1(x1))))))	(87)
1(0(5(0(x1))))	→	1(5(2(0(2(0(x1))))))	(60)
1(2(5(0(x1))))	→	1(5(2(2(0(x1)))))	(62)
1(0(1(2(0(x1)))))	→	1(1(0(2(0(0(x1))))))	(76)
1(4(0(3(0(x1)))))	→	3(4(4(1(0(0(x1))))))	(80)
1(5(2(5(0(x1)))))	→	1(5(5(0(2(0(x1))))))	(86)
1(0(3(5(2(x1)))))	→	1(3(2(2(0(5(x1))))))	(89)
1(0(1(2(5(x1)))))	→	1(5(1(3(2(0(x1))))))	(92)
1(0(3(2(5(x1)))))	→	3(2(2(0(5(1(x1))))))	(93)
1(4(0(3(5(x1)))))	→	1(3(2(0(5(4(x1))))))	(94)
5(0(5(0(x1))))	→	5(5(0(2(0(x1)))))	(61)
5(2(5(0(x1))))	→	5(5(2(2(0(x1)))))	(63)
5(0(3(2(0(x1)))))	→	3(5(0(0(2(0(x1))))))	(78)
4(0(1(2(x1))))	→	2(2(2(0(4(1(x1))))))	(66)
4(0(1(0(0(x1)))))	→	4(1(0(2(0(0(x1))))))	(68)
4(0(2(3(0(x1)))))	→	4(3(2(0(0(4(x1))))))	(81)
3(2(1(0(x1))))	→	3(1(0(2(0(x1)))))	(51)
3(2(1(0(x1))))	→	1(3(3(2(0(x1)))))	(52)
3(0(1(2(x1))))	→	1(3(2(2(0(4(x1))))))	(65)
3(0(2(1(0(x1)))))	→	3(1(4(0(2(0(x1))))))	(71)
3(2(3(1(0(x1)))))	→	1(3(3(2(0(0(x1))))))	(74)
3(1(0(3(0(x1)))))	→	3(1(3(4(0(0(x1))))))	(79)
3(2(5(4(0(x1)))))	→	5(4(3(2(2(0(x1))))))	(82)
3(0(0(5(0(x1)))))	→	5(0(3(0(2(0(x1))))))	(83)
3(0(0(5(2(x1)))))	→	3(5(0(0(2(0(x1))))))	(88)

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

3^#(1(0(3(0(x1)))))

→

3^#(4(0(0(x1))))

(206)

could be deleted.

1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the arctic semiring over the integers

[4(x₁)]	=	-∞ · x₁ + 0
[3^#(x₁)]	=	0 · x₁ + 0
[1(x₁)]	=	0 · x₁ + 0
[3(x₁)]	=	-∞ · x₁ + 0
[5(x₁)]	=	-∞ · x₁ + 0
[2(x₁)]	=	1 · x₁ + 0
[0(x₁)]	=	-∞ · x₁ + 4

together with the usable rules

1(0(5(0(x1))))	→	1(5(2(0(2(0(x1))))))	(60)
1(2(5(0(x1))))	→	1(5(2(2(0(x1)))))	(62)
1(0(1(2(0(x1)))))	→	1(1(0(2(0(0(x1))))))	(76)
1(4(0(3(0(x1)))))	→	3(4(4(1(0(0(x1))))))	(80)
1(5(2(5(0(x1)))))	→	1(5(5(0(2(0(x1))))))	(86)
1(0(3(5(2(x1)))))	→	1(3(2(2(0(5(x1))))))	(89)
1(0(1(2(5(x1)))))	→	1(5(1(3(2(0(x1))))))	(92)
1(0(3(2(5(x1)))))	→	3(2(2(0(5(1(x1))))))	(93)
1(4(0(3(5(x1)))))	→	1(3(2(0(5(4(x1))))))	(94)
5(0(5(0(x1))))	→	5(5(0(2(0(x1)))))	(61)
5(2(5(0(x1))))	→	5(5(2(2(0(x1)))))	(63)
5(0(3(2(0(x1)))))	→	3(5(0(0(2(0(x1))))))	(78)
3(2(1(0(x1))))	→	3(1(0(2(0(x1)))))	(51)
3(2(1(0(x1))))	→	1(3(3(2(0(x1)))))	(52)
3(0(1(2(x1))))	→	1(3(2(2(0(4(x1))))))	(65)
3(0(2(1(0(x1)))))	→	3(1(4(0(2(0(x1))))))	(71)
3(2(3(1(0(x1)))))	→	1(3(3(2(0(0(x1))))))	(74)
3(1(0(3(0(x1)))))	→	3(1(3(4(0(0(x1))))))	(79)
3(2(5(4(0(x1)))))	→	5(4(3(2(2(0(x1))))))	(82)
3(0(0(5(0(x1)))))	→	5(0(3(0(2(0(x1))))))	(83)
3(0(0(5(2(x1)))))	→	3(5(0(0(2(0(x1))))))	(88)

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

3^#(1(0(3(0(x1)))))

→

3^#(1(3(4(0(0(x1))))))

(208)

could be deleted.

1.1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

4^#(0(0(1(2(x1)))))	→	4^#(4(1(x1)))	(238)
4^#(0(0(1(2(x1)))))	→	1^#(x1)	(236)
1^#(4(0(3(5(x1)))))	→	4^#(x1)	(266)
4^#(0(2(3(0(x1)))))	→	4^#(x1)	(213)
4^#(0(1(2(x1))))	→	1^#(x1)	(155)
1^#(0(3(2(5(x1)))))	→	1^#(x1)	(261)

1.1.1.2 Reduction Pair Processor with Usable Rules

Using the

prec(4^#)	=	stat(4^#)	=	lex
prec(1^#)	=	stat(1^#)	=	lex
prec(5)	=	stat(5)	=	lex
prec(4)	=	stat(4)	=	lex
prec(3)	=	stat(3)	=	lex
prec(0)	=	stat(0)	=	lex
prec(1)	=	stat(1)	=	lex
prec(2)	=	stat(2)	=	lex

π(4^#)	=	1
π(1^#)	=	1
π(5)	=	1
π(4)	=	1
π(3)	=	1
π(0)	=	1
π(1)	=	[1]
π(2)	=	1

together with the usable rules

1(0(5(0(x1))))	→	1(5(2(0(2(0(x1))))))	(60)
1(2(5(0(x1))))	→	1(5(2(2(0(x1)))))	(62)
1(0(1(2(0(x1)))))	→	1(1(0(2(0(0(x1))))))	(76)
1(4(0(3(0(x1)))))	→	3(4(4(1(0(0(x1))))))	(80)
1(5(2(5(0(x1)))))	→	1(5(5(0(2(0(x1))))))	(86)
1(0(3(5(2(x1)))))	→	1(3(2(2(0(5(x1))))))	(89)
1(0(1(2(5(x1)))))	→	1(5(1(3(2(0(x1))))))	(92)
1(0(3(2(5(x1)))))	→	3(2(2(0(5(1(x1))))))	(93)
1(4(0(3(5(x1)))))	→	1(3(2(0(5(4(x1))))))	(94)
5(0(5(0(x1))))	→	5(5(0(2(0(x1)))))	(61)
5(2(5(0(x1))))	→	5(5(2(2(0(x1)))))	(63)
5(0(3(2(0(x1)))))	→	3(5(0(0(2(0(x1))))))	(78)
4(0(1(2(x1))))	→	2(2(2(0(4(1(x1))))))	(66)
4(0(1(0(0(x1)))))	→	4(1(0(2(0(0(x1))))))	(68)
4(0(2(3(0(x1)))))	→	4(3(2(0(0(4(x1))))))	(81)
4(0(0(1(2(x1)))))	→	2(0(0(4(4(1(x1))))))	(87)

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

4^#(0(0(1(2(x1)))))	→	1^#(x1)	(236)
4^#(0(1(2(x1))))	→	1^#(x1)	(155)

could be deleted.

1.1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
1^#(0(3(2(5(x1))))) → 1^#(x1) (261)

1.1.1.2.1.1 Size-Change Termination

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

1^#(0(3(2(5(x1))))) → 1^#(x1) (261)

1 > 1

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

The 2^nd component contains the pair

4^#(0(0(1(2(x1)))))	→	4^#(4(1(x1)))	(238)
4^#(0(2(3(0(x1)))))	→	4^#(x1)	(213)

1.1.1.2.1.2 Reduction Pair Processor with Usable Rules

Using the

prec(4^#)	=	0	stat(4^#)	=	lex
prec(5)	=	0	stat(5)	=	lex
prec(4)	=	0	stat(4)	=	lex
prec(3)	=	0	stat(3)	=	lex
prec(0)	=	0	stat(0)	=	lex
prec(1)	=	1	stat(1)	=	lex
prec(2)	=	0	stat(2)	=	lex

π(4^#)	=	1
π(5)	=	[]
π(4)	=	[]
π(3)	=	1
π(0)	=	1
π(1)	=	[]
π(2)	=	1

together with the usable rules

4(0(1(2(x1))))	→	2(2(2(0(4(1(x1))))))	(66)
4(0(1(0(0(x1)))))	→	4(1(0(2(0(0(x1))))))	(68)
4(0(2(3(0(x1)))))	→	4(3(2(0(0(4(x1))))))	(81)
4(0(0(1(2(x1)))))	→	2(0(0(4(4(1(x1))))))	(87)

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

4^#(0(0(1(2(x1)))))

→

4^#(4(1(x1)))

(238)

could be deleted.

1.1.1.2.1.2.1 Size-Change Termination

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

4^#(0(2(3(0(x1)))))	→	4^#(x1)	(213)

1	>	1

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

Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212774)

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

1.1.1.1 Reduction Pair Processor with Usable Rules

1.1.1.1.1 Reduction Pair Processor with Usable Rules

1.1.1.1.1.1 P is empty

1.1.1.2 Reduction Pair Processor with Usable Rules

1.1.1.2.1 Dependency Graph Processor

1.1.1.2.1.1 Size-Change Termination

1.1.1.2.1.2 Reduction Pair Processor with Usable Rules

1.1.1.2.1.2.1 Size-Change Termination