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

(0) Obligation:

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


The TRS R consists of the following rules:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

Rewrite Strategy: FULL

(1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
del(.([], .(y, z)))
del(.(x, .([], z)))

The defined contexts are:
f([], x1, x2, x3)

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

(2) 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:

del(.(x, .(y, z))) → f(=(x, y), x, y, z)
f(true, x, y, z) → del(.(y, z))
f(false, x, y, z) → .(x, del(.(y, z)))
=(nil, nil) → true
=(.(x, y), nil) → false
=(nil, .(y, z)) → false
=(.(x, y), .(u, v)) → and(=(x, u), =(y, v))

Rewrite Strategy: INNERMOST

(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))

Tuples:

DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
='(nil, nil) → c3
='(.(z0, z1), nil) → c4
='(nil, .(z0, z1)) → c5
='(.(z0, z1), .(u, v)) → c6(='(z0, u), ='(z1, v))

S tuples:

DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
='(nil, nil) → c3
='(.(z0, z1), nil) → c4
='(nil, .(z0, z1)) → c5
='(.(z0, z1), .(u, v)) → c6(='(z0, u), ='(z1, v))

K tuples:none
Defined Rule Symbols:

del, f, =

Defined Pair Symbols:

DEL, F, ='

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

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

Removed 4 trailing nodes:

='(nil, nil) → c3
='(nil, .(z0, z1)) → c5
='(.(z0, z1), nil) → c4
='(.(z0, z1), .(u, v)) → c6(='(z0, u), ='(z1, v))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))

Tuples:

DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))

S tuples:

DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2), ='(z0, z1))
F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))

K tuples:none
Defined Rule Symbols:

del, f, =

Defined Pair Symbols:

DEL, F

Compound Symbols:

c, c1, c2

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

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))
=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))

Tuples:

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

S tuples:

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

K tuples:none
Defined Rule Symbols:

del, f, =

Defined Pair Symbols:

F, DEL

Compound Symbols:

c1, c2, c

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

del(.(z0, .(z1, z2))) → f(=(z0, z1), z0, z1, z2)
f(true, z0, z1, z2) → del(.(z1, z2))
f(false, z0, z1, z2) → .(z0, del(.(z1, z2)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))

Tuples:

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

S tuples:

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

K tuples:none
Defined Rule Symbols:

=

Defined Pair Symbols:

F, DEL

Compound Symbols:

c1, c2, c

(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.

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

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

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

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

POL(.(x₁, x₂)) = [2] + x₁ + x₂   
POL(=(x₁, x₂)) = 0   
POL(DEL(x₁)) = x₁   
POL(F(x₁, x₂, x₃, x₄)) = [3] + x₃ + x₄   
POL(and(x₁, x₂)) = 0   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(false) = 0   
POL(nil) = 0   
POL(true) = 0   
POL(u) = 0   
POL(v) = 0   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

=(nil, nil) → true
=(.(z0, z1), nil) → false
=(nil, .(z0, z1)) → false
=(.(z0, z1), .(u, v)) → and(=(z0, u), =(z1, v))

Tuples:

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

S tuples:none
K tuples:

F(true, z0, z1, z2) → c1(DEL(.(z1, z2)))
F(false, z0, z1, z2) → c2(DEL(.(z1, z2)))
DEL(.(z0, .(z1, z2))) → c(F(=(z0, z1), z0, z1, z2))

Defined Rule Symbols:

=

Defined Pair Symbols:

F, DEL

Compound Symbols:

c1, c2, c

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

The set S is empty

(14) BOUNDS(1, 1)