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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 7 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)), 71 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:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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:
not0(0) → 0
or0(0, 0) → 0
implies0(0, 0) → 1
or1(0, 0) → 1
or1(0, 0) → 2
implies1(0, 2) → 1
implies1(0, 0) → 3
or1(0, 3) → 1
or1(0, 2) → 1
or1(0, 0) → 3
implies1(0, 2) → 3
or1(0, 3) → 3
implies2(0, 0) → 4
or2(0, 4) → 1
or1(0, 2) → 3
or2(0, 4) → 3
or1(0, 0) → 4
implies1(0, 2) → 4
or1(0, 3) → 4
or1(0, 2) → 4
or2(0, 4) → 4

(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:

implies(not(z0), z1) → or(z0, z1)
implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))
Tuples:

IMPLIES(not(z0), z1) → c
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:

IMPLIES(not(z0), z1) → c
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
K tuples:none
Defined Rule Symbols:

implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

IMPLIES(not(z0), z1) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

implies(not(z0), z1) → or(z0, z1)
implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))
Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
K tuples:none
Defined Rule Symbols:

implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

implies(not(z0), z1) → or(z0, z1)
implies(not(z0), or(z1, z2)) → implies(z1, or(z0, z2))
implies(z0, or(z1, z2)) → or(z1, implies(z0, z2))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, c2

(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.

IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(IMPLIES(x1, x2)) = [2]x2   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(not(x1)) = 0   
POL(or(x1, x2)) = [2] + x2   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
K tuples:

IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, 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.

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
We considered the (Usable) Rules:none
And the Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(IMPLIES(x1, x2)) = x1 + x2   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(not(x1)) = [1] + x1   
POL(or(x1, x2)) = x1 + x2   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
S tuples:none
K tuples:

IMPLIES(z0, or(z1, z2)) → c2(IMPLIES(z0, z2))
IMPLIES(not(z0), or(z1, z2)) → c1(IMPLIES(z1, or(z0, z2)))
Defined Rule Symbols:none

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c1, c2

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

The set S is empty

(14) BOUNDS(1, 1)