The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

not(tt) → ff
not(ff) → tt
or(tt, x) → tt
or(ff, x) → x
eq(0, 0) → tt
eq(s(x), 0) → ff
eq(0, s(y)) → ff
eq(s(x), s(y)) → eq(x, y)
main(phi) → ver(phi, nil)
in(x, nil) → ff
in(x, cons(a, l)) → or(eq(x, a), in(x, l))
ver(Var(x), t) → in(x, t)
ver(Or(phi, psi), t) → or(ver(phi, t), ver(psi, t))
ver(Not(phi), t) → not(ver(phi, t))
ver(Exists(n, phi), t) → or(ver(phi, cons(n, t)), ver(phi, t))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
MAIN(z0) → c8(VER(z0, nil))
IN(z0, cons(z1, z2)) → c10(OR(eq(z0, z1), in(z0, z2)), EQ(z0, z1), IN(z0, z2))
VER(Var(z0), t) → c11(IN(z0, t))
VER(Or(z0, z1), t) → c12(OR(ver(z0, t), ver(z1, t)), VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(NOT(ver(z0, t)), VER(z0, t))
VER(Exists(z0, z1), t) → c14(OR(ver(z1, cons(z0, t)), ver(z1, t)), VER(z1, cons(z0, t)), VER(z1, t))

S tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
MAIN(z0) → c8(VER(z0, nil))
IN(z0, cons(z1, z2)) → c10(OR(eq(z0, z1), in(z0, z2)), EQ(z0, z1), IN(z0, z2))
VER(Var(z0), t) → c11(IN(z0, t))
VER(Or(z0, z1), t) → c12(OR(ver(z0, t), ver(z1, t)), VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(NOT(ver(z0, t)), VER(z0, t))
VER(Exists(z0, z1), t) → c14(OR(ver(z1, cons(z0, t)), ver(z1, t)), VER(z1, cons(z0, t)), VER(z1, t))

K tuples:none
Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, MAIN, IN, VER

Compound Symbols:

c7, c8, c10, c11, c12, c13, c14

(3) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)

Removed 2 of 7 dangling nodes:

MAIN(z0) → c8(VER(z0, nil))
VER(Var(z0), t) → c11(IN(z0, t))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(OR(eq(z0, z1), in(z0, z2)), EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(OR(ver(z0, t), ver(z1, t)), VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(NOT(ver(z0, t)), VER(z0, t))
VER(Exists(z0, z1), t) → c14(OR(ver(z1, cons(z0, t)), ver(z1, t)), VER(z1, cons(z0, t)), VER(z1, t))

S tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(OR(eq(z0, z1), in(z0, z2)), EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(OR(ver(z0, t), ver(z1, t)), VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(NOT(ver(z0, t)), VER(z0, t))
VER(Exists(z0, z1), t) → c14(OR(ver(z1, cons(z0, t)), ver(z1, t)), VER(z1, cons(z0, t)), VER(z1, t))

K tuples:none
Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

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

Removed 5 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

S tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

K tuples:none
Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))

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

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

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

POL(EQ(x₁, x₂)) = [3]   
POL(Exists(x₁, x₂)) = 0   
POL(IN(x₁, x₂)) = [4]x₂   
POL(Not(x₁)) = [2]   
POL(Or(x₁, x₂)) = [3]   
POL(VER(x₁, x₂)) = 0   
POL(c10(x₁, x₂)) = x₁ + x₂   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [2] + x₂   
POL(s(x₁)) = [3]   
POL(t) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

S tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

K tuples:

IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))

Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))

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

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

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

POL(EQ(x₁, x₂)) = [1] + x₂ + x₁·x₂   
POL(Exists(x₁, x₂)) = 0   
POL(IN(x₁, x₂)) = x₂ + x₁·x₂   
POL(Not(x₁)) = 0   
POL(Or(x₁, x₂)) = 0   
POL(VER(x₁, x₂)) = 0   
POL(c10(x₁, x₂)) = x₁ + x₂   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [1] + x₁ + x₂   
POL(s(x₁)) = [1] + x₁   
POL(t) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

S tuples:

VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

K tuples:

IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
EQ(s(z0), s(z1)) → c7(EQ(z0, z1))

Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

VER(Not(z0), t) → c13(VER(z0, t))

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

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

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

POL(EQ(x₁, x₂)) = [2] + [2]x₂   
POL(Exists(x₁, x₂)) = x₂   
POL(IN(x₁, x₂)) = [4]x₂   
POL(Not(x₁)) = [4] + x₁   
POL(Or(x₁, x₂)) = x₁ + x₂   
POL(VER(x₁, x₂)) = [2]x₁   
POL(c10(x₁, x₂)) = x₁ + x₂   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [4] + x₁ + x₂   
POL(s(x₁)) = x₁   
POL(t) = [2]   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

S tuples:

VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

K tuples:

IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
VER(Not(z0), t) → c13(VER(z0, t))

Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

VER(Exists(z0, z1), t) → c14(VER(z1, t))

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

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

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

POL(EQ(x₁, x₂)) = [3]   
POL(Exists(x₁, x₂)) = [2] + x₂   
POL(IN(x₁, x₂)) = x₂   
POL(Not(x₁)) = x₁   
POL(Or(x₁, x₂)) = x₁ + x₂   
POL(VER(x₁, x₂)) = [2]x₁   
POL(c10(x₁, x₂)) = x₁ + x₂   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [4] + x₂   
POL(s(x₁)) = 0   
POL(t) = [2]   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

S tuples:

VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))

K tuples:

IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))

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

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

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

POL(EQ(x₁, x₂)) = [3]   
POL(Exists(x₁, x₂)) = x₂   
POL(IN(x₁, x₂)) = [4]x₂   
POL(Not(x₁)) = x₁   
POL(Or(x₁, x₂)) = [4] + x₁ + x₂   
POL(VER(x₁, x₂)) = [1] + [4]x₁ + [2]x₂   
POL(c10(x₁, x₂)) = x₁ + x₂   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c13(x₁)) = x₁   
POL(c14(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(cons(x₁, x₂)) = [1] + x₁ + x₂   
POL(s(x₁)) = 0   
POL(t) = [1]   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

not(tt) → ff
not(ff) → tt
or(tt, z0) → tt
or(ff, z0) → z0
eq(0, 0) → tt
eq(s(z0), 0) → ff
eq(0, s(z0)) → ff
eq(s(z0), s(z1)) → eq(z0, z1)
main(z0) → ver(z0, nil)
in(z0, nil) → ff
in(z0, cons(z1, z2)) → or(eq(z0, z1), in(z0, z2))
ver(Var(z0), t) → in(z0, t)
ver(Or(z0, z1), t) → or(ver(z0, t), ver(z1, t))
ver(Not(z0), t) → not(ver(z0, t))
ver(Exists(z0, z1), t) → or(ver(z1, cons(z0, t)), ver(z1, t))

Tuples:

EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))

S tuples:none
K tuples:

IN(z0, cons(z1, z2)) → c10(EQ(z0, z1), IN(z0, z2))
EQ(s(z0), s(z1)) → c7(EQ(z0, z1))
VER(Not(z0), t) → c13(VER(z0, t))
VER(Exists(z0, z1), t) → c14(VER(z1, t))
VER(Or(z0, z1), t) → c12(VER(z0, t), VER(z1, t))

Defined Rule Symbols:

not, or, eq, main, in, ver

Defined Pair Symbols:

EQ, IN, VER

Compound Symbols:

c7, c10, c12, c13, c14

(17) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(18) BOUNDS(O(1), O(1))