(0) Obligation:

Runtime Complexity TRS:
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) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) 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

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

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

REVAPP(Nil, z0) → c1

(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)))
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

(5) 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)

(6) 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

(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(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)) = [5] + x1 + x2   
POL(REVAPP(x1, x2)) = [4]x1 + [3]x2   
POL(c(x1)) = x1   

(8) 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

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

The set S is empty

(10) BOUNDS(O(1), O(1))