Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212892)

The rewrite relation of the following TRS is considered.

0(0(x1))	→	1(0(2(0(2(x1)))))	(1)
0(0(x1))	→	0(2(3(4(0(2(x1))))))	(2)
0(3(x1))	→	0(2(3(2(x1))))	(3)
0(3(x1))	→	2(0(2(1(3(x1)))))	(4)
0(3(x1))	→	2(3(0(2(2(x1)))))	(5)
0(3(x1))	→	0(2(2(2(3(2(x1))))))	(6)
0(0(0(x1)))	→	0(2(0(0(x1))))	(7)
0(0(4(x1)))	→	2(4(0(0(3(2(x1))))))	(8)
0(0(5(x1)))	→	0(5(2(0(x1))))	(9)
0(1(3(x1)))	→	3(2(2(1(0(x1)))))	(10)
0(3(0(x1)))	→	2(0(2(3(4(0(x1))))))	(11)
0(4(3(x1)))	→	3(0(2(4(x1))))	(12)
0(4(3(x1)))	→	1(0(2(1(3(4(x1))))))	(13)
0(4(3(x1)))	→	2(2(4(3(4(0(x1))))))	(14)
0(4(5(x1)))	→	2(4(0(5(2(5(x1))))))	(15)
0(4(5(x1)))	→	5(3(2(4(4(0(x1))))))	(16)
0(5(0(x1)))	→	0(5(2(4(0(x1)))))	(17)
0(5(0(x1)))	→	5(0(2(3(2(0(x1))))))	(18)
0(5(3(x1)))	→	3(2(5(3(2(0(x1))))))	(19)
1(0(0(x1)))	→	1(0(2(5(0(2(x1))))))	(20)
1(0(3(x1)))	→	2(1(3(0(5(2(x1))))))	(21)
1(0(4(x1)))	→	3(0(2(1(4(x1)))))	(22)
1(0(4(x1)))	→	2(1(2(1(4(0(x1))))))	(23)
3(0(0(x1)))	→	3(0(2(0(x1))))	(24)
3(0(3(x1)))	→	3(0(2(3(x1))))	(25)
3(0(3(x1)))	→	3(3(2(2(0(x1)))))	(26)
3(0(4(x1)))	→	2(2(3(4(0(x1)))))	(27)
3(0(4(x1)))	→	2(3(4(2(0(3(x1))))))	(28)
3(0(4(x1)))	→	5(5(2(4(0(3(x1))))))	(29)
3(3(4(x1)))	→	3(2(1(3(2(4(x1))))))	(30)
3(5(0(x1)))	→	3(5(1(0(2(1(x1))))))	(31)
4(0(4(x1)))	→	0(2(4(4(0(x1)))))	(32)
4(0(4(x1)))	→	1(4(0(2(4(0(x1))))))	(33)
0(0(4(3(x1))))	→	3(0(0(2(1(4(x1))))))	(34)
0(0(5(3(x1))))	→	0(5(2(5(0(3(x1))))))	(35)
0(1(0(3(x1))))	→	2(1(0(3(0(x1)))))	(36)
0(1(1(0(x1))))	→	2(1(0(0(1(2(x1))))))	(37)
0(4(0(0(x1))))	→	0(4(0(2(1(0(x1))))))	(38)
0(4(1(3(x1))))	→	2(1(4(2(0(3(x1))))))	(39)
0(4(5(3(x1))))	→	4(3(2(5(0(x1)))))	(40)
0(4(5(3(x1))))	→	4(3(5(0(2(4(x1))))))	(41)
0(5(0(4(x1))))	→	5(0(0(2(4(x1)))))	(42)
0(5(5(3(x1))))	→	2(3(5(5(0(x1)))))	(43)
1(3(0(4(x1))))	→	4(0(1(2(3(x1)))))	(44)
4(0(3(5(x1))))	→	5(2(3(4(2(0(x1))))))	(45)
0(0(1(0(4(x1)))))	→	0(0(0(1(4(2(x1))))))	(46)
0(0(5(5(3(x1)))))	→	0(5(0(3(2(5(x1))))))	(47)
3(0(4(3(4(x1)))))	→	3(0(3(4(2(4(x1))))))	(48)
3(3(0(4(5(x1)))))	→	1(4(5(3(0(3(x1))))))	(49)
3(4(2(0(3(x1)))))	→	2(3(0(2(3(4(x1))))))	(50)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

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

0(0(x1))	→	2(0(2(0(1(x1)))))	(51)
0(0(x1))	→	2(0(4(3(2(0(x1))))))	(52)
3(0(x1))	→	2(3(2(0(x1))))	(53)
3(0(x1))	→	3(1(2(0(2(x1)))))	(54)
3(0(x1))	→	2(2(0(3(2(x1)))))	(55)
3(0(x1))	→	2(3(2(2(2(0(x1))))))	(56)
0(0(0(x1)))	→	0(0(2(0(x1))))	(57)
4(0(0(x1)))	→	2(3(0(0(4(2(x1))))))	(58)
5(0(0(x1)))	→	0(2(5(0(x1))))	(59)
3(1(0(x1)))	→	0(1(2(2(3(x1)))))	(60)
0(3(0(x1)))	→	0(4(3(2(0(2(x1))))))	(61)
3(4(0(x1)))	→	4(2(0(3(x1))))	(62)
3(4(0(x1)))	→	4(3(1(2(0(1(x1))))))	(63)
3(4(0(x1)))	→	0(4(3(4(2(2(x1))))))	(64)
5(4(0(x1)))	→	5(2(5(0(4(2(x1))))))	(65)
5(4(0(x1)))	→	0(4(4(2(3(5(x1))))))	(66)
0(5(0(x1)))	→	0(4(2(5(0(x1)))))	(67)
0(5(0(x1)))	→	0(2(3(2(0(5(x1))))))	(68)
3(5(0(x1)))	→	0(2(3(5(2(3(x1))))))	(69)
0(0(1(x1)))	→	2(0(5(2(0(1(x1))))))	(70)
3(0(1(x1)))	→	2(5(0(3(1(2(x1))))))	(71)
4(0(1(x1)))	→	4(1(2(0(3(x1)))))	(72)
4(0(1(x1)))	→	0(4(1(2(1(2(x1))))))	(73)
0(0(3(x1)))	→	0(2(0(3(x1))))	(74)
3(0(3(x1)))	→	3(2(0(3(x1))))	(75)
3(0(3(x1)))	→	0(2(2(3(3(x1)))))	(76)
4(0(3(x1)))	→	0(4(3(2(2(x1)))))	(77)
4(0(3(x1)))	→	3(0(2(4(3(2(x1))))))	(78)
4(0(3(x1)))	→	3(0(4(2(5(5(x1))))))	(79)
4(3(3(x1)))	→	4(2(3(1(2(3(x1))))))	(80)
0(5(3(x1)))	→	1(2(0(1(5(3(x1))))))	(81)
4(0(4(x1)))	→	0(4(4(2(0(x1)))))	(82)
4(0(4(x1)))	→	0(4(2(0(4(1(x1))))))	(83)
3(4(0(0(x1))))	→	4(1(2(0(0(3(x1))))))	(84)
3(5(0(0(x1))))	→	3(0(5(2(5(0(x1))))))	(85)
3(0(1(0(x1))))	→	0(3(0(1(2(x1)))))	(86)
0(1(1(0(x1))))	→	2(1(0(0(1(2(x1))))))	(37)
0(0(4(0(x1))))	→	0(1(2(0(4(0(x1))))))	(87)
3(1(4(0(x1))))	→	3(0(2(4(1(2(x1))))))	(88)
3(5(4(0(x1))))	→	0(5(2(3(4(x1)))))	(89)
3(5(4(0(x1))))	→	4(2(0(5(3(4(x1))))))	(90)
4(0(5(0(x1))))	→	4(2(0(0(5(x1)))))	(91)
3(5(5(0(x1))))	→	0(5(5(3(2(x1)))))	(92)
4(0(3(1(x1))))	→	3(2(1(0(4(x1)))))	(93)
5(3(0(4(x1))))	→	0(2(4(3(2(5(x1))))))	(94)
4(0(1(0(0(x1)))))	→	2(4(1(0(0(0(x1))))))	(95)
3(5(5(0(0(x1)))))	→	5(2(3(0(5(0(x1))))))	(96)
4(3(4(0(3(x1)))))	→	4(2(4(3(0(3(x1))))))	(97)
5(4(0(3(3(x1)))))	→	3(0(3(5(4(1(x1))))))	(98)
3(0(2(4(3(x1)))))	→	4(3(2(0(3(2(x1))))))	(99)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 148 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

0^#(5(0(x1)))	→	0^#(5(x1))	(148)
0^#(5(0(x1)))	→	5^#(x1)	(149)
5^#(0(0(x1)))	→	5^#(0(x1))	(118)
5^#(4(0(x1)))	→	3^#(5(x1))	(142)
3^#(1(0(x1)))	→	3^#(x1)	(120)
3^#(4(0(x1)))	→	0^#(3(x1))	(126)
3^#(4(0(x1)))	→	3^#(x1)	(127)
3^#(5(0(x1)))	→	3^#(x1)	(153)
3^#(0(3(x1)))	→	3^#(3(x1))	(167)
3^#(4(0(0(x1))))	→	0^#(0(3(x1)))	(192)
3^#(4(0(0(x1))))	→	0^#(3(x1))	(193)
3^#(4(0(0(x1))))	→	3^#(x1)	(194)
3^#(5(0(0(x1))))	→	5^#(0(x1))	(198)
5^#(4(0(x1)))	→	5^#(x1)	(143)
5^#(3(0(4(x1))))	→	5^#(x1)	(229)
5^#(4(0(3(3(x1)))))	→	3^#(0(3(5(4(1(x1))))))	(239)
3^#(0(1(0(x1))))	→	0^#(3(0(1(2(x1)))))	(199)
3^#(5(4(0(x1))))	→	3^#(4(x1))	(210)
3^#(5(4(0(x1))))	→	4^#(x1)	(211)
4^#(0(0(x1)))	→	3^#(0(0(4(2(x1)))))	(113)
3^#(5(4(0(x1))))	→	0^#(5(3(4(x1))))	(213)
3^#(5(4(0(x1))))	→	5^#(3(4(x1)))	(214)
3^#(5(5(0(0(x1)))))	→	3^#(0(5(0(x1))))	(233)
3^#(5(5(0(0(x1)))))	→	0^#(5(0(x1)))	(234)
3^#(5(5(0(0(x1)))))	→	5^#(0(x1))	(235)
4^#(0(1(x1)))	→	0^#(3(x1))	(160)
4^#(0(1(x1)))	→	3^#(x1)	(161)
4^#(0(3(x1)))	→	5^#(5(x1))	(178)
4^#(0(3(x1)))	→	5^#(x1)	(179)
4^#(0(4(x1)))	→	0^#(x1)	(186)
4^#(0(5(0(x1))))	→	0^#(0(5(x1)))	(216)
4^#(0(5(0(x1))))	→	0^#(5(x1))	(217)
4^#(0(5(0(x1))))	→	5^#(x1)	(218)
4^#(0(3(1(x1))))	→	0^#(4(x1))	(224)
4^#(0(3(1(x1))))	→	4^#(x1)	(225)
4^#(0(1(0(0(x1)))))	→	0^#(0(0(x1)))	(231)
4^#(3(4(0(3(x1)))))	→	4^#(3(0(3(x1))))	(237)
4^#(3(4(0(3(x1)))))	→	3^#(0(3(x1)))	(238)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

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

the pairs

0^#(5(0(x1)))	→	0^#(5(x1))	(148)
0^#(5(0(x1)))	→	5^#(x1)	(149)
5^#(0(0(x1)))	→	5^#(0(x1))	(118)
5^#(4(0(x1)))	→	3^#(5(x1))	(142)
3^#(1(0(x1)))	→	3^#(x1)	(120)
3^#(4(0(x1)))	→	0^#(3(x1))	(126)
3^#(4(0(x1)))	→	3^#(x1)	(127)
3^#(5(0(x1)))	→	3^#(x1)	(153)
3^#(0(3(x1)))	→	3^#(3(x1))	(167)
3^#(4(0(0(x1))))	→	0^#(0(3(x1)))	(192)
3^#(4(0(0(x1))))	→	0^#(3(x1))	(193)
3^#(4(0(0(x1))))	→	3^#(x1)	(194)
3^#(5(0(0(x1))))	→	5^#(0(x1))	(198)
5^#(4(0(x1)))	→	5^#(x1)	(143)
5^#(3(0(4(x1))))	→	5^#(x1)	(229)
3^#(0(1(0(x1))))	→	0^#(3(0(1(2(x1)))))	(199)
3^#(5(4(0(x1))))	→	3^#(4(x1))	(210)
4^#(0(0(x1)))	→	3^#(0(0(4(2(x1)))))	(113)
3^#(5(4(0(x1))))	→	0^#(5(3(4(x1))))	(213)
3^#(5(4(0(x1))))	→	5^#(3(4(x1)))	(214)
3^#(5(5(0(0(x1)))))	→	0^#(5(0(x1)))	(234)
3^#(5(5(0(0(x1)))))	→	5^#(0(x1))	(235)
4^#(0(1(x1)))	→	0^#(3(x1))	(160)
4^#(0(1(x1)))	→	3^#(x1)	(161)
4^#(0(3(x1)))	→	5^#(5(x1))	(178)
4^#(0(3(x1)))	→	5^#(x1)	(179)
4^#(0(4(x1)))	→	0^#(x1)	(186)
4^#(0(5(0(x1))))	→	0^#(0(5(x1)))	(216)
4^#(0(5(0(x1))))	→	0^#(5(x1))	(217)
4^#(0(5(0(x1))))	→	5^#(x1)	(218)
4^#(0(3(1(x1))))	→	0^#(4(x1))	(224)
4^#(0(3(1(x1))))	→	4^#(x1)	(225)
4^#(0(1(0(0(x1)))))	→	0^#(0(0(x1)))	(231)
4^#(3(4(0(3(x1)))))	→	3^#(0(3(x1)))	(238)

could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

3^#(5(5(0(0(x1)))))

→

3^#(0(5(0(x1))))

(233)

1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

0(0(x1))	→	2(0(2(0(1(x1)))))	(51)
0(0(x1))	→	2(0(4(3(2(0(x1))))))	(52)
0(0(0(x1)))	→	0(0(2(0(x1))))	(57)
0(3(0(x1)))	→	0(4(3(2(0(2(x1))))))	(61)
0(5(0(x1)))	→	0(4(2(5(0(x1)))))	(67)
0(5(0(x1)))	→	0(2(3(2(0(5(x1))))))	(68)
0(0(1(x1)))	→	2(0(5(2(0(1(x1))))))	(70)
0(0(3(x1)))	→	0(2(0(3(x1))))	(74)
0(5(3(x1)))	→	1(2(0(1(5(3(x1))))))	(81)
0(1(1(0(x1))))	→	2(1(0(0(1(2(x1))))))	(37)
0(0(4(0(x1))))	→	0(1(2(0(4(0(x1))))))	(87)

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

3^#(5(5(0(0(x1)))))

→

3^#(0(5(0(x1))))

(233)

could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

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

→

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

(237)

1.1.1.1.1.2 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

together with the usable rules

3(0(x1))	→	2(3(2(0(x1))))	(53)
3(0(x1))	→	3(1(2(0(2(x1)))))	(54)
3(0(x1))	→	2(2(0(3(2(x1)))))	(55)
3(0(x1))	→	2(3(2(2(2(0(x1))))))	(56)
3(1(0(x1)))	→	0(1(2(2(3(x1)))))	(60)
3(4(0(x1)))	→	4(2(0(3(x1))))	(62)
3(4(0(x1)))	→	4(3(1(2(0(1(x1))))))	(63)
3(4(0(x1)))	→	0(4(3(4(2(2(x1))))))	(64)
3(5(0(x1)))	→	0(2(3(5(2(3(x1))))))	(69)
3(0(1(x1)))	→	2(5(0(3(1(2(x1))))))	(71)
3(0(3(x1)))	→	3(2(0(3(x1))))	(75)
3(0(3(x1)))	→	0(2(2(3(3(x1)))))	(76)
3(4(0(0(x1))))	→	4(1(2(0(0(3(x1))))))	(84)
3(5(0(0(x1))))	→	3(0(5(2(5(0(x1))))))	(85)
3(0(1(0(x1))))	→	0(3(0(1(2(x1)))))	(86)
3(1(4(0(x1))))	→	3(0(2(4(1(2(x1))))))	(88)
3(5(4(0(x1))))	→	0(5(2(3(4(x1)))))	(89)
3(5(4(0(x1))))	→	4(2(0(5(3(4(x1))))))	(90)
3(5(5(0(x1))))	→	0(5(5(3(2(x1)))))	(92)
3(5(5(0(0(x1)))))	→	5(2(3(0(5(0(x1))))))	(96)
3(0(2(4(3(x1)))))	→	4(3(2(0(3(2(x1))))))	(99)
0(0(x1))	→	2(0(2(0(1(x1)))))	(51)
0(0(x1))	→	2(0(4(3(2(0(x1))))))	(52)
0(0(0(x1)))	→	0(0(2(0(x1))))	(57)
0(3(0(x1)))	→	0(4(3(2(0(2(x1))))))	(61)
0(5(0(x1)))	→	0(4(2(5(0(x1)))))	(67)
0(5(0(x1)))	→	0(2(3(2(0(5(x1))))))	(68)
0(0(1(x1)))	→	2(0(5(2(0(1(x1))))))	(70)
0(0(3(x1)))	→	0(2(0(3(x1))))	(74)
0(5(3(x1)))	→	1(2(0(1(5(3(x1))))))	(81)
0(1(1(0(x1))))	→	2(1(0(0(1(2(x1))))))	(37)
0(0(4(0(x1))))	→	0(1(2(0(4(0(x1))))))	(87)

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

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

→

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

(237)

could be deleted.

1.1.1.1.1.2.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

0^#(0(x1)) → 0^#(1(x1)) (101)

0^#(1(1(0(x1)))) → 0^#(0(1(2(x1)))) (202)

1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[0(x₁)] = 1 · x₁

[1(x₁)] = 1 · x₁

[2(x₁)] = 1 · x₁

[0^#(x₁)] = 1 · x₁

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
1.1.1.2.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[0^#(x₁)] = 1 · x₁

[0(x₁)] = 1 + 1 · x₁

[1(x₁)] = 1 + 1 · x₁

[2(x₁)] = 0

the pair
0^#(1(1(0(x1)))) → 0^#(0(1(2(x1)))) (202)
could be deleted.
1.1.1.2.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[0^#(x₁)] = 1 · x₁

[0(x₁)] = 1 + 1 · x₁

[1(x₁)] = 1 · x₁

the pair
0^#(0(x1)) → 0^#(1(x1)) (101)
could be deleted.
1.1.1.2.1.1.1 P is empty

There are no pairs anymore.

0^#(0(x1))	→	0^#(1(x1))	(101)
0^#(1(1(0(x1))))	→	0^#(0(1(2(x1))))	(202)

Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/212892)

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

1.1.1.1 Reduction Pair Processor

1.1.1.1.1 Dependency Graph Processor

1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

1.1.1.1.1.1.1 P is empty

1.1.1.1.1.2 Reduction Pair Processor with Usable Rules

1.1.1.1.1.2.1 P is empty

1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.2.1 Reduction Pair Processor

1.1.1.2.1.1 Reduction Pair Processor

1.1.1.2.1.1.1 P is empty