(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
from(X) → cons(X, n__from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
activate(n__from(X)) → from(X)
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
SQUARE(z0) → c13(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(z0))
S tuples:
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
SQUARE(z0) → c13(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(z0))
K tuples:none
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
2NDSPOS, 2NDSNEG, PI, PLUS, TIMES, SQUARE, ACTIVATE
Compound Symbols:
c3, c4, c6, c7, c8, c10, c12, c13, c14
(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
SQUARE(z0) → c13(TIMES(z0, z0))
Removed 1 trailing nodes:
ACTIVATE(n__from(z0)) → c14(FROM(z0))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
S tuples:
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
K tuples:none
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
2NDSPOS, 2NDSNEG, PI, PLUS, TIMES
Compound Symbols:
c3, c4, c6, c7, c8, c10, c12
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 5 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
S tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
K tuples:none
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI
Compound Symbols:
c10, c12, c3, c4, c6, c7, c8
(7) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
PI(z0) → c8(2NDSPOS(z0, from(0)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
S tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
K tuples:
PI(z0) → c8(2NDSPOS(z0, from(0)))
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI
Compound Symbols:
c10, c12, c3, c4, c6, c7, c8
(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
We considered the (Usable) Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [4]
POL(2NDSNEG(x1, x2)) = 0
POL(2NDSPOS(x1, x2)) = 0
POL(PI(x1)) = [2] + [2]x1
POL(PLUS(x1, x2)) = 0
POL(TIMES(x1, x2)) = [2]x1
POL(activate(x1)) = 0
POL(c10(x1)) = x1
POL(c12(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(cons(x1, x2)) = 0
POL(cons2(x1, x2)) = 0
POL(from(x1)) = [5] + x1
POL(n__from(x1)) = x1
POL(plus(x1, x2)) = [5]
POL(s(x1)) = [4] + x1
POL(times(x1, x2)) = 0
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
S tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
K tuples:
PI(z0) → c8(2NDSPOS(z0, from(0)))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI
Compound Symbols:
c10, c12, c3, c4, c6, c7, c8
(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
We considered the (Usable) Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [2]
POL(2NDSNEG(x1, x2)) = [1] + [2]x1
POL(2NDSPOS(x1, x2)) = [4] + [2]x1
POL(PI(x1)) = [4] + [4]x1
POL(PLUS(x1, x2)) = 0
POL(TIMES(x1, x2)) = 0
POL(activate(x1)) = 0
POL(c10(x1)) = x1
POL(c12(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(cons(x1, x2)) = 0
POL(cons2(x1, x2)) = 0
POL(from(x1)) = [1] + [3]x1
POL(n__from(x1)) = 0
POL(plus(x1, x2)) = [3] + [2]x1 + x2
POL(s(x1)) = [2] + x1
POL(times(x1, x2)) = [3] + x1 + [3]x2
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
S tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
K tuples:
PI(z0) → c8(2NDSPOS(z0, from(0)))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI
Compound Symbols:
c10, c12, c3, c4, c6, c7, c8
(13) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
S tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
K tuples:
PI(z0) → c8(2NDSPOS(z0, from(0)))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI
Compound Symbols:
c10, c12, c3, c4, c6, c7, c8
(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
We considered the (Usable) Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
And the Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(2NDSNEG(x1, x2)) = [1]
POL(2NDSPOS(x1, x2)) = [1]
POL(PI(x1)) = [2] + x1 + [3]x12
POL(PLUS(x1, x2)) = [2]x1
POL(TIMES(x1, x2)) = x1·x2
POL(activate(x1)) = [2] + x1
POL(c10(x1)) = x1
POL(c12(x1, x2)) = x1 + x2
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(cons(x1, x2)) = 0
POL(cons2(x1, x2)) = 0
POL(from(x1)) = [2]
POL(n__from(x1)) = [2]
POL(plus(x1, x2)) = 0
POL(s(x1)) = [2] + x1
POL(times(x1, x2)) = 0
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
PI(z0) → c8(2NDSPOS(z0, from(0)))
S tuples:none
K tuples:
PI(z0) → c8(2NDSPOS(z0, from(0)))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
Defined Rule Symbols:
from, 2ndspos, 2ndsneg, pi, plus, times, square, activate
Defined Pair Symbols:
PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI
Compound Symbols:
c10, c12, c3, c4, c6, c7, c8
(17) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(18) BOUNDS(O(1), O(1))