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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 17 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 87 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 0 ms)
        ↳12 CdtProblem
        ↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳14 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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, 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 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
00() → 0
s0(0) → 0
+0(0, 0) → 1
-0(0, 0) → 2
+1(0, 0) → 3
s1(3) → 1
01() → 2
-1(0, 0) → 2
s1(3) → 3
0 → 1
0 → 2
0 → 3

(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), z1) → s(+(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
-'(0, z0) → c2
-'(z0, 0) → c3
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
-'(0, z0) → c2
-'(z0, 0) → c3
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 3 trailing nodes:

+'(0, z0) → c
-'(0, z0) → c2
-'(z0, 0) → c3

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, 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.

-'(s(z0), s(z1)) → c4(-'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = 0   
POL(-'(x1, x2)) = x1 + x2   
POL(c1(x1)) = x1   
POL(c4(x1)) = x1   
POL(s(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
K tuples:

-'(s(z0), s(z1)) → c4(-'(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

(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), z1) → c1(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = x1   
POL(-'(x1, x2)) = 0   
POL(c1(x1)) = x1   
POL(c4(x1)) = x1   
POL(s(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:none
K tuples:

-'(s(z0), s(z1)) → c4(-'(z0, z1))
+'(s(z0), z1) → c1(+'(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

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

The set S is empty

(14) BOUNDS(1, 1)