Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/Rubio_04/polo2)

The rewrite relation of the following TRS is considered.

dx(X)	→	one	(1)
dx(a)	→	zero	(2)
dx(plus(ALPHA,BETA))	→	plus(dx(ALPHA),dx(BETA))	(3)
dx(times(ALPHA,BETA))	→	plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA)))	(4)
dx(minus(ALPHA,BETA))	→	minus(dx(ALPHA),dx(BETA))	(5)
dx(neg(ALPHA))	→	neg(dx(ALPHA))	(6)
dx(div(ALPHA,BETA))	→	minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two))))	(7)
dx(ln(ALPHA))	→	div(dx(ALPHA),ALPHA)	(8)
dx(exp(ALPHA,BETA))	→	plus(times(BETA,times(exp(ALPHA,minus(BETA,one)),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA))))	(9)

The evaluation strategy is innermost.

Property / Task

Determine bounds on the runtime complexity.

Answer / Result

An upperbound for the complexity is O(n).

Proof (by AProVE @ termCOMP 2023)

1 Dependency Tuples

We get the following set of dependency tuples:

dx^#(z0)

→

(11)

originates from

dx(z0)

→

one

(10)

dx^#(a)

→

(12)

originates from

dx(a)

→

zero

(2)

dx^#(plus(z0,z1))

→

c2(dx^#(z0),dx^#(z1))

(14)

originates from

dx(plus(z0,z1))

→

plus(dx(z0),dx(z1))

(13)

dx^#(times(z0,z1))

→

c3(dx^#(z0),dx^#(z1))

(16)

originates from

dx(times(z0,z1))

→

plus(times(z1,dx(z0)),times(z0,dx(z1)))

(15)

dx^#(minus(z0,z1))

→

c4(dx^#(z0),dx^#(z1))

(18)

originates from

dx(minus(z0,z1))

→

minus(dx(z0),dx(z1))

(17)

dx^#(neg(z0))

→

c5(dx^#(z0))

(20)

originates from

dx(neg(z0))

→

neg(dx(z0))

(19)

dx^#(div(z0,z1))

→

c6(dx^#(z0),dx^#(z1))

(22)

originates from

dx(div(z0,z1))

→

minus(div(dx(z0),z1),times(z0,div(dx(z1),exp(z1,two))))

(21)

dx^#(ln(z0))

→

c7(dx^#(z0))

(24)

originates from

dx(ln(z0))

→

div(dx(z0),z0)

(23)

dx^#(exp(z0,z1))

→

c8(dx^#(z0),dx^#(z1))

(26)

originates from

dx(exp(z0,z1))

→

plus(times(z1,times(exp(z0,minus(z1,one)),dx(z0))),times(exp(z0,z1),times(ln(z0),dx(z1))))

(25)

Moreover, we add the following terms to the innermost strategy.

dx^#(z0)

dx^#(a)

dx^#(plus(z0,z1))

dx^#(times(z0,z1))

dx^#(minus(z0,z1))

dx^#(neg(z0))

dx^#(div(z0,z1))

dx^#(ln(z0))

dx^#(exp(z0,z1))

1.1 Usable Rules

We remove the following rules since they are not usable.

dx(z0)	→	one	(10)
dx(a)	→	zero	(2)
dx(plus(z0,z1))	→	plus(dx(z0),dx(z1))	(13)
dx(times(z0,z1))	→	plus(times(z1,dx(z0)),times(z0,dx(z1)))	(15)
dx(minus(z0,z1))	→	minus(dx(z0),dx(z1))	(17)
dx(neg(z0))	→	neg(dx(z0))	(19)
dx(div(z0,z1))	→	minus(div(dx(z0),z1),times(z0,div(dx(z1),exp(z1,two))))	(21)
dx(ln(z0))	→	div(dx(z0),z0)	(23)
dx(exp(z0,z1))	→	plus(times(z1,times(exp(z0,minus(z1,one)),dx(z0))),times(exp(z0,z1),times(ln(z0),dx(z1))))	(25)

1.1.1 Rule Shifting

The rules

dx^#(a)	→	c1	(12)
dx^#(times(z0,z1))	→	c3(dx^#(z0),dx^#(z1))	(16)
dx^#(minus(z0,z1))	→	c4(dx^#(z0),dx^#(z1))	(18)
dx^#(neg(z0))	→	c5(dx^#(z0))	(20)
dx^#(div(z0,z1))	→	c6(dx^#(z0),dx^#(z1))	(22)
dx^#(ln(z0))	→	c7(dx^#(z0))	(24)
dx^#(exp(z0,z1))	→	c8(dx^#(z0),dx^#(z1))	(26)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c]	=	0
[c1]	=	0
[c2(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c3(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c5(x₁)]	=	1 · x₁ + 0
[c6(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c7(x₁)]	=	1 · x₁ + 0
[c8(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[dx^#(x₁)]	=	3 · x₁ + 0
[a]	=	1
[plus(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[times(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂
[minus(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂
[neg(x₁)]	=	3 + 1 · x₁
[div(x₁, x₂)]	=	3 + 1 · x₁ + 1 · x₂
[ln(x₁)]	=	3 + 1 · x₁
[exp(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂

which has the intended complexity. Here, only the following usable rules have been considered:

dx^#(z0)	→	c	(11)
dx^#(a)	→	c1	(12)
dx^#(plus(z0,z1))	→	c2(dx^#(z0),dx^#(z1))	(14)
dx^#(times(z0,z1))	→	c3(dx^#(z0),dx^#(z1))	(16)
dx^#(minus(z0,z1))	→	c4(dx^#(z0),dx^#(z1))	(18)
dx^#(neg(z0))	→	c5(dx^#(z0))	(20)
dx^#(div(z0,z1))	→	c6(dx^#(z0),dx^#(z1))	(22)
dx^#(ln(z0))	→	c7(dx^#(z0))	(24)
dx^#(exp(z0,z1))	→	c8(dx^#(z0),dx^#(z1))	(26)

1.1.1.1 Rule Shifting

The rules

dx^#(z0)

→

(11)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c]	=	0
[c1]	=	0
[c2(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c3(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c5(x₁)]	=	1 · x₁ + 0
[c6(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c7(x₁)]	=	1 · x₁ + 0
[c8(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[dx^#(x₁)]	=	1 + 1 · x₁
[a]	=	0
[plus(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[times(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[minus(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[neg(x₁)]	=	1 · x₁ + 0
[div(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[ln(x₁)]	=	1 · x₁ + 0
[exp(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂

which has the intended complexity. Here, only the following usable rules have been considered:

dx^#(z0)	→	c	(11)
dx^#(a)	→	c1	(12)
dx^#(plus(z0,z1))	→	c2(dx^#(z0),dx^#(z1))	(14)
dx^#(times(z0,z1))	→	c3(dx^#(z0),dx^#(z1))	(16)
dx^#(minus(z0,z1))	→	c4(dx^#(z0),dx^#(z1))	(18)
dx^#(neg(z0))	→	c5(dx^#(z0))	(20)
dx^#(div(z0,z1))	→	c6(dx^#(z0),dx^#(z1))	(22)
dx^#(ln(z0))	→	c7(dx^#(z0))	(24)
dx^#(exp(z0,z1))	→	c8(dx^#(z0),dx^#(z1))	(26)

1.1.1.1.1 Rule Shifting

The rules

dx^#(plus(z0,z1))

→

c2(dx^#(z0),dx^#(z1))

(14)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c]	=	0
[c1]	=	0
[c2(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c3(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c5(x₁)]	=	1 · x₁ + 0
[c6(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c7(x₁)]	=	1 · x₁ + 0
[c8(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[dx^#(x₁)]	=	1 · x₁ + 0
[a]	=	0
[plus(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[times(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[minus(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[neg(x₁)]	=	1 · x₁ + 0
[div(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[ln(x₁)]	=	1 · x₁ + 0
[exp(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂

which has the intended complexity. Here, only the following usable rules have been considered:

dx^#(z0)	→	c	(11)
dx^#(a)	→	c1	(12)
dx^#(plus(z0,z1))	→	c2(dx^#(z0),dx^#(z1))	(14)
dx^#(times(z0,z1))	→	c3(dx^#(z0),dx^#(z1))	(16)
dx^#(minus(z0,z1))	→	c4(dx^#(z0),dx^#(z1))	(18)
dx^#(neg(z0))	→	c5(dx^#(z0))	(20)
dx^#(div(z0,z1))	→	c6(dx^#(z0),dx^#(z1))	(22)
dx^#(ln(z0))	→	c7(dx^#(z0))	(24)
dx^#(exp(z0,z1))	→	c8(dx^#(z0),dx^#(z1))	(26)

1.1.1.1.1.1 R is empty

There are no rules in the TRS R. Hence, R/S has complexity O(1).