(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
terms(N) → cons(recip(sqr(N)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(Y)
Rewrite Strategy: INNERMOST
(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3, 4, 5]
transitions:
cons0(0) → 0
recip0(0) → 0
00() → 0
s0() → 0
nil0() → 0
terms0(0) → 1
sqr0(0) → 2
dbl0(0) → 3
add0(0, 0) → 4
first0(0, 0) → 5
sqr1(0) → 7
recip1(7) → 6
cons1(6) → 1
01() → 2
s1() → 2
01() → 3
s1() → 3
s1() → 4
nil1() → 5
cons1(0) → 5
01() → 7
s1() → 7
0 → 4
(2) BOUNDS(1, n^1)
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
terms(z0) → cons(recip(sqr(z0)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, z0) → z0
add(s, z0) → s
first(0, z0) → nil
first(s, cons(z0)) → cons(z0)
Tuples:
TERMS(z0) → c(SQR(z0))
SQR(0) → c1
SQR(s) → c2
DBL(0) → c3
DBL(s) → c4
ADD(0, z0) → c5
ADD(s, z0) → c6
FIRST(0, z0) → c7
FIRST(s, cons(z0)) → c8
S tuples:
TERMS(z0) → c(SQR(z0))
SQR(0) → c1
SQR(s) → c2
DBL(0) → c3
DBL(s) → c4
ADD(0, z0) → c5
ADD(s, z0) → c6
FIRST(0, z0) → c7
FIRST(s, cons(z0)) → c8
K tuples:none
Defined Rule Symbols:
terms, sqr, dbl, add, first
Defined Pair Symbols:
TERMS, SQR, DBL, ADD, FIRST
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 9 trailing nodes:
SQR(s) → c2
FIRST(s, cons(z0)) → c8
ADD(s, z0) → c6
FIRST(0, z0) → c7
DBL(s) → c4
DBL(0) → c3
ADD(0, z0) → c5
SQR(0) → c1
TERMS(z0) → c(SQR(z0))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
terms(z0) → cons(recip(sqr(z0)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, z0) → z0
add(s, z0) → s
first(0, z0) → nil
first(s, cons(z0)) → cons(z0)
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:
terms, sqr, dbl, add, first
Defined Pair Symbols:none
Compound Symbols:none
(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(8) BOUNDS(1, 1)