WORST_CASE(?,O(n^2))
proof of input_gduykWz0Ze.trs
# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished


The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
(3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
(4) CdtProblem
(5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
(6) CdtProblem
(7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 71 ms]
(8) CdtProblem
(9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 42 ms]
(10) CdtProblem
(11) CdtKnowledgeProof [FINISHED, 0 ms]
(12) BOUNDS(1, 1)


----------------------------------------

(0)
Obligation:
The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   first(0, Z) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   sel(0, cons(X, Z)) -> X
   sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
   from(X) -> n__from(X)
   first(X1, X2) -> n__first(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(X) -> X

S is empty.
Rewrite Strategy: PARALLEL_INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   first(0, z0) -> nil
   first(s(z0), cons(z1, z2)) -> cons(z1, n__first(z0, activate(z2)))
   first(z0, z1) -> n__first(z0, z1)
   sel(0, cons(z0, z1)) -> z0
   sel(s(z0), cons(z1, z2)) -> sel(z0, activate(z2))
   activate(n__from(z0)) -> from(z0)
   activate(n__first(z0, z1)) -> first(z0, z1)
   activate(z0) -> z0
Tuples:
   FROM(z0) -> c
   FROM(z0) -> c1
   FIRST(0, z0) -> c2
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   FIRST(z0, z1) -> c4
   SEL(0, cons(z0, z1)) -> c5
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__from(z0)) -> c7(FROM(z0))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
   ACTIVATE(z0) -> c9
S tuples:
   FROM(z0) -> c
   FROM(z0) -> c1
   FIRST(0, z0) -> c2
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   FIRST(z0, z1) -> c4
   SEL(0, cons(z0, z1)) -> c5
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__from(z0)) -> c7(FROM(z0))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
   ACTIVATE(z0) -> c9
K tuples:none
Defined Rule Symbols:   from_1, first_2, sel_2, activate_1

Defined Pair Symbols:   FROM_1, FIRST_2, SEL_2, ACTIVATE_1

Compound Symbols:   c, c1, c2, c3_1, c4, c5, c6_2, c7_1, c8_1, c9


----------------------------------------

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 7 trailing nodes:
   FIRST(z0, z1) -> c4
   ACTIVATE(n__from(z0)) -> c7(FROM(z0))
   SEL(0, cons(z0, z1)) -> c5
   FROM(z0) -> c1
   FROM(z0) -> c
   FIRST(0, z0) -> c2
   ACTIVATE(z0) -> c9

----------------------------------------

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   first(0, z0) -> nil
   first(s(z0), cons(z1, z2)) -> cons(z1, n__first(z0, activate(z2)))
   first(z0, z1) -> n__first(z0, z1)
   sel(0, cons(z0, z1)) -> z0
   sel(s(z0), cons(z1, z2)) -> sel(z0, activate(z2))
   activate(n__from(z0)) -> from(z0)
   activate(n__first(z0, z1)) -> first(z0, z1)
   activate(z0) -> z0
Tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
S tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
K tuples:none
Defined Rule Symbols:   from_1, first_2, sel_2, activate_1

Defined Pair Symbols:   FIRST_2, SEL_2, ACTIVATE_1

Compound Symbols:   c3_1, c6_2, c8_1


----------------------------------------

(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   sel(0, cons(z0, z1)) -> z0
   sel(s(z0), cons(z1, z2)) -> sel(z0, activate(z2))

----------------------------------------

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   activate(n__from(z0)) -> from(z0)
   activate(n__first(z0, z1)) -> first(z0, z1)
   activate(z0) -> z0
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   first(0, z0) -> nil
   first(s(z0), cons(z1, z2)) -> cons(z1, n__first(z0, activate(z2)))
   first(z0, z1) -> n__first(z0, z1)
Tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
S tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
K tuples:none
Defined Rule Symbols:   activate_1, from_1, first_2

Defined Pair Symbols:   FIRST_2, SEL_2, ACTIVATE_1

Compound Symbols:   c3_1, c6_2, c8_1


----------------------------------------

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
We considered the (Usable) Rules:none
And the Tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [3]
   POL(ACTIVATE(x_1)) = 0
   POL(FIRST(x_1, x_2)) = 0
   POL(SEL(x_1, x_2)) = [2]x_1
   POL(activate(x_1)) = [3]
   POL(c3(x_1)) = x_1
   POL(c6(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1
   POL(first(x_1, x_2)) = [3]
   POL(from(x_1)) = [3] + [3]x_1
   POL(n__first(x_1, x_2)) = [3] + x_1
   POL(n__from(x_1)) = [3] + x_1
   POL(nil) = [3]
   POL(s(x_1)) = [2] + x_1

----------------------------------------

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   activate(n__from(z0)) -> from(z0)
   activate(n__first(z0, z1)) -> first(z0, z1)
   activate(z0) -> z0
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   first(0, z0) -> nil
   first(s(z0), cons(z1, z2)) -> cons(z1, n__first(z0, activate(z2)))
   first(z0, z1) -> n__first(z0, z1)
Tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
S tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
K tuples:
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
Defined Rule Symbols:   activate_1, from_1, first_2

Defined Pair Symbols:   FIRST_2, SEL_2, ACTIVATE_1

Compound Symbols:   c3_1, c6_2, c8_1


----------------------------------------

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
We considered the (Usable) Rules:
   first(z0, z1) -> n__first(z0, z1)
   from(z0) -> cons(z0, n__from(s(z0)))
   first(s(z0), cons(z1, z2)) -> cons(z1, n__first(z0, activate(z2)))
   first(0, z0) -> nil
   from(z0) -> n__from(z0)
   activate(n__from(z0)) -> from(z0)
   activate(z0) -> z0
   activate(n__first(z0, z1)) -> first(z0, z1)
And the Tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [2]
   POL(ACTIVATE(x_1)) = x_1
   POL(FIRST(x_1, x_2)) = x_1 + x_2
   POL(SEL(x_1, x_2)) = x_1*x_2 + x_1^2
   POL(activate(x_1)) = [1] + x_1
   POL(c3(x_1)) = x_1
   POL(c6(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_2
   POL(first(x_1, x_2)) = [1] + x_1 + x_2
   POL(from(x_1)) = 0
   POL(n__first(x_1, x_2)) = x_1 + x_2
   POL(n__from(x_1)) = 0
   POL(nil) = [2]
   POL(s(x_1)) = [1] + x_1

----------------------------------------

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   activate(n__from(z0)) -> from(z0)
   activate(n__first(z0, z1)) -> first(z0, z1)
   activate(z0) -> z0
   from(z0) -> cons(z0, n__from(s(z0)))
   from(z0) -> n__from(z0)
   first(0, z0) -> nil
   first(s(z0), cons(z1, z2)) -> cons(z1, n__first(z0, activate(z2)))
   first(z0, z1) -> n__first(z0, z1)
Tuples:
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
S tuples:
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
K tuples:
   SEL(s(z0), cons(z1, z2)) -> c6(SEL(z0, activate(z2)), ACTIVATE(z2))
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
Defined Rule Symbols:   activate_1, from_1, first_2

Defined Pair Symbols:   FIRST_2, SEL_2, ACTIVATE_1

Compound Symbols:   c3_1, c6_2, c8_1


----------------------------------------

(11) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   ACTIVATE(n__first(z0, z1)) -> c8(FIRST(z0, z1))
   FIRST(s(z0), cons(z1, z2)) -> c3(ACTIVATE(z2))
Now S is empty
----------------------------------------

(12)
BOUNDS(1, 1)