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

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y) → x
if(false, x, y) → y
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if(le(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, cons(y, z)) → if(eq(x, y), z, cons(y, del(x, z)))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
min(x, cons(y, z)) →⁺ if(le(x, y), min(x, z), min(y, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [z / cons(y, z)].
The result substitution is [ ].

The rewrite sequence
min(x, cons(y, z)) →⁺ if(le(x, y), min(x, z), min(y, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [2].
The pumping substitution is [z / cons(y, z)].
The result substitution is [x / y].

(2) BOUNDS(2^n, INF)