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:

+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))

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 3.

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) → 1
s1(0) → 1
s1(0) → 3
01() → 5
+1(0, 5) → 4
+1(3, 4) → 2
s1(2) → 1
s1(0) → 4
s2(0) → 2
s2(0) → 7
02() → 9
+2(0, 9) → 8
+2(7, 8) → 6
s2(6) → 2
s1(0) → 8
s3(0) → 6
s2(6) → 6
0 → 1
5 → 4
9 → 8

(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:

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

Tuples:

+'(0, z0) → c
+'(s(z0), 0) → c1
+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)), +'(z1, 0))

S tuples:

+'(0, z0) → c
+'(s(z0), 0) → c1
+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)), +'(z1, 0))

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

+'(0, z0) → c
+'(s(z0), 0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)), +'(z1, 0))

S tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)), +'(z1, 0))

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

S tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), 0) → s(z0)

Tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

S tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

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

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

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

We considered the (Usable) Rules:

+(s(z0), 0) → s(z0)
+(0, z0) → z0

And the Tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

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

POL(+(x₁, x₂)) = x₁ + x₂   
POL(+'(x₁, x₂)) = x₂   
POL(0) = 0   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), 0) → s(z0)

Tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

S tuples:none
K tuples:

+'(s(z0), s(z1)) → c2(+'(s(z0), +(z1, 0)))

Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c2

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

The set S is empty

(14) BOUNDS(1, 1)