The Runtime Complexity (full) 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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

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:

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(<=(x, y), min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(=(x, y), z, .(y, del(x, z)))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
if/0
<=/0
<=/1
=/0
=/1

(4) Obligation:

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

msort(nil) → nil
msort(.(x, y)) → .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) → x
min(x, .(y, z)) → if(min(x, z), min(y, z))
del(x, nil) → nil
del(x, .(y, z)) → if(z, .(y, del(x, z)))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
min(x, .(y, z)) →⁺ if(min(x, z), min(y, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [z / .(y, z)].
The result substitution is [ ].

The rewrite sequence
min(x, .(y, z)) →⁺ if(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 / .(y, z)].
The result substitution is [x / y].

(6) BOUNDS(2^n, INF)