The Runtime Complexity (full) 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:

circ(cons(a, s), t) → cons(msubst(a, t), circ(s, t))
circ(cons(lift, s), cons(a, t)) → cons(a, circ(s, t))
circ(cons(lift, s), cons(lift, t)) → cons(lift, circ(s, t))
circ(circ(s, t), u) → circ(s, circ(t, u))
circ(s, id) → s
circ(id, s) → s
circ(cons(lift, s), circ(cons(lift, t), u)) → circ(cons(lift, circ(s, t)), u)
subst(a, id) → a
msubst(a, id) → a
msubst(msubst(a, s), t) → msubst(a, circ(s, t))

Rewrite Strategy: FULL

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

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

circ(cons(a, s), t) → cons(msubst(a, t), circ(s, t))
circ(cons(lift, s), cons(a, t)) → cons(a, circ(s, t))
circ(cons(lift, s), cons(lift, t)) → cons(lift, circ(s, t))
circ(circ(s, t), u) → circ(s, circ(t, u))
circ(s, id) → s
circ(id, s) → s
circ(cons(lift, s), circ(cons(lift, t), u)) → circ(cons(lift, circ(s, t)), u)
subst(a, id) → a
msubst(a, id) → a
msubst(msubst(a, s), t) → msubst(a, circ(s, t))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
circ(cons(a, s), t) →⁺ cons(msubst(a, t), circ(s, t))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [s / cons(a, s)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)