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

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(x, s(0))) → +(s(s(0)), f(x))
f(+(x, y)) → *(f(x), f(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 2.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1]
transitions:
00() → 0
s0(0) → 0
*0(0, 0) → 0
+0(0, 0) → 0
f0(0) → 1
01() → 2
s1(2) → 1
s1(2) → 3
s1(3) → 1
s1(3) → 4
01() → 6
f1(6) → 5
*1(4, 5) → 1
s1(3) → 7
f1(0) → 8
+1(7, 8) → 1
f1(0) → 9
f1(0) → 10
*1(9, 10) → 1
s1(2) → 8
s1(2) → 9
s1(2) → 10
02() → 11
s2(11) → 5
s1(3) → 8
s1(3) → 9
s1(3) → 10
*1(4, 5) → 8
*1(4, 5) → 9
*1(4, 5) → 10
+1(7, 8) → 8
+1(7, 8) → 9
+1(7, 8) → 10
*1(9, 10) → 8
*1(9, 10) → 9
*1(9, 10) → 10

(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:

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(z0, s(0))) → +(s(s(0)), f(z0))
f(+(z0, z1)) → *(f(z0), f(z1))

Tuples:

F(0) → c
F(s(0)) → c1
F(s(0)) → c2(F(0))
F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:

F(0) → c
F(s(0)) → c1
F(s(0)) → c2(F(0))
F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1, c2, c3, c4

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing nodes:

F(s(0)) → c2(F(0))
F(s(0)) → c1
F(0) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(z0, s(0))) → +(s(s(0)), f(z0))
f(+(z0, z1)) → *(f(z0), f(z1))

Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0) → s(0)
f(s(0)) → s(s(0))
f(s(0)) → *(s(s(0)), f(0))
f(+(z0, s(0))) → +(s(s(0)), f(z0))
f(+(z0, z1)) → *(f(z0), f(z1))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

(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.

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

We considered the (Usable) Rules:none
And the Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x₁, x₂)) = [1] + x₁ + x₂   
POL(0) = 0   
POL(F(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = [1]   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:none
K tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

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

The set S is empty

(12) BOUNDS(1, 1)