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

(0) Obligation:

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

intersect'ii'in(cons(X, X0), cons(X, X1)) → intersect'ii'out
intersect'ii'in(Xs, cons(X0, Ys)) → u'1'1(intersect'ii'in(Xs, Ys))
u'1'1(intersect'ii'out) → intersect'ii'out
intersect'ii'in(cons(X0, Xs), Ys) → u'2'1(intersect'ii'in(Xs, Ys))
u'2'1(intersect'ii'out) → intersect'ii'out
reduce'ii'in(sequent(cons(if(A, B), Fs), Gs), NF) → u'3'1(reduce'ii'in(sequent(cons(x'2b(x'2d(B), A), Fs), Gs), NF))
u'3'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(cons(iff(A, B), Fs), Gs), NF) → u'4'1(reduce'ii'in(sequent(cons(x'2a(if(A, B), if(B, A)), Fs), Gs), NF))
u'4'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(cons(x'2a(F1, F2), Fs), Gs), NF) → u'5'1(reduce'ii'in(sequent(cons(F1, cons(F2, Fs)), Gs), NF))
u'5'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(cons(x'2b(F1, F2), Fs), Gs), NF) → u'6'1(reduce'ii'in(sequent(cons(F1, Fs), Gs), NF), F2, Fs, Gs, NF)
u'6'1(reduce'ii'out, F2, Fs, Gs, NF) → u'6'2(reduce'ii'in(sequent(cons(F2, Fs), Gs), NF))
u'6'2(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(cons(x'2d(F1), Fs), Gs), NF) → u'7'1(reduce'ii'in(sequent(Fs, cons(F1, Gs)), NF))
u'7'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(Fs, cons(if(A, B), Gs)), NF) → u'8'1(reduce'ii'in(sequent(Fs, cons(x'2b(x'2d(B), A), Gs)), NF))
u'8'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(Fs, cons(iff(A, B), Gs)), NF) → u'9'1(reduce'ii'in(sequent(Fs, cons(x'2a(if(A, B), if(B, A)), Gs)), NF))
u'9'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(cons(p(V), Fs), Gs), sequent(Left, Right)) → u'10'1(reduce'ii'in(sequent(Fs, Gs), sequent(cons(p(V), Left), Right)))
u'10'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(Fs, cons(x'2b(G1, G2), Gs)), NF) → u'11'1(reduce'ii'in(sequent(Fs, cons(G1, cons(G2, Gs))), NF))
u'11'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(Fs, cons(x'2a(G1, G2), Gs)), NF) → u'12'1(reduce'ii'in(sequent(Fs, cons(G1, Gs)), NF), Fs, G2, Gs, NF)
u'12'1(reduce'ii'out, Fs, G2, Gs, NF) → u'12'2(reduce'ii'in(sequent(Fs, cons(G2, Gs)), NF))
u'12'2(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(Fs, cons(x'2d(G1), Gs)), NF) → u'13'1(reduce'ii'in(sequent(cons(G1, Fs), Gs), NF))
u'13'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(nil, cons(p(V), Gs)), sequent(Left, Right)) → u'14'1(reduce'ii'in(sequent(nil, Gs), sequent(Left, cons(p(V), Right))))
u'14'1(reduce'ii'out) → reduce'ii'out
reduce'ii'in(sequent(nil, nil), sequent(F1, F2)) → u'15'1(intersect'ii'in(F1, F2))
u'15'1(intersect'ii'out) → reduce'ii'out
tautology'i'in(F) → u'16'1(reduce'ii'in(sequent(nil, cons(F, nil)), sequent(nil, nil)))
u'16'1(reduce'ii'out) → tautology'i'out

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
intersect'ii'in(Xs, cons(X0, Ys)) →⁺ u'1'1(intersect'ii'in(Xs, Ys))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [Ys / cons(X0, Ys)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)