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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 11 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 63 ms)
        ↳10 CdtProblem
        ↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳12 BOUNDS(1, 1)

(0) Obligation:

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


The TRS R consists of the following rules:

revapp(Cons(x, xs), rest) → revapp(xs, Cons(x, rest))
revapp(Nil, rest) → rest
goal(xs, ys) → revapp(xs, ys)

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]
transitions:
Cons0(0, 0) → 0
Nil0() → 0
revapp0(0, 0) → 1
goal0(0, 0) → 2
Cons1(0, 0) → 3
revapp1(0, 3) → 1
revapp1(0, 0) → 2
Cons1(0, 3) → 3
revapp1(0, 3) → 2
0 → 1
0 → 2
3 → 1
3 → 2

(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:

revapp(Cons(z0, z1), z2) → revapp(z1, Cons(z0, z2))
revapp(Nil, z0) → z0
goal(z0, z1) → revapp(z0, z1)
Tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
REVAPP(Nil, z0) → c1
GOAL(z0, z1) → c2(REVAPP(z0, z1))
S tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
REVAPP(Nil, z0) → c1
GOAL(z0, z1) → c2(REVAPP(z0, z1))
K tuples:none
Defined Rule Symbols:

revapp, goal

Defined Pair Symbols:

REVAPP, GOAL

Compound Symbols:

c, c1, c2

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c2(REVAPP(z0, z1))
Removed 1 trailing nodes:

REVAPP(Nil, z0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

revapp(Cons(z0, z1), z2) → revapp(z1, Cons(z0, z2))
revapp(Nil, z0) → z0
goal(z0, z1) → revapp(z0, z1)
Tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
S tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
K tuples:none
Defined Rule Symbols:

revapp, goal

Defined Pair Symbols:

REVAPP

Compound Symbols:

c

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

revapp(Cons(z0, z1), z2) → revapp(z1, Cons(z0, z2))
revapp(Nil, z0) → z0
goal(z0, z1) → revapp(z0, z1)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
S tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

REVAPP

Compound Symbols:

c

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
We considered the (Usable) Rules:none
And the Tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(Cons(x1, x2)) = [3] + x2   
POL(REVAPP(x1, x2)) = [3]x1 + x2   
POL(c(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
S tuples:none
K tuples:

REVAPP(Cons(z0, z1), z2) → c(REVAPP(z1, Cons(z0, z2)))
Defined Rule Symbols:none

Defined Pair Symbols:

REVAPP

Compound Symbols:

c

(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(12) BOUNDS(1, 1)