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$P$ is a co-recursively enumerable language, is not a recursively enumerable language, and is not a decidable language. The stipulated order on halting times includes $\infty \ge \infty$.

First, failure has a finite witness. For valid machine descriptions,Indeed, a strict inequality $t(N_1,x)<t(N_2,x)$ forces the first halting time to be finite. The second can be either a larger finite time or infinity. Whether the displayed condition holds for fixed $x,t$ is decidable by a finite simulation, including the initial configurations when $t=0$.

Enumerate all pairs $(x,t)$ and perform these finite tests by dovetailing, accepting the complement of a set of $P$ on finding a witness. Equivalently, at stage $s$, inspect all inputs of length at most $s$ through $s$ transitions. A genuine violation eventually appears; if there is no violation, the procedure never accepts. Malformed descriptions or differing input alphabets can be recognized and accepted as nonmembers immediately. Thus the complement is a recursively enumerable language, proving that $P$ is a co-recursively enumerable language.

For non-enumerability, use the standard result that the complement of a set of the halting problemis not a recursively enumerable language. Given $⟨M,w⟩$, effectively construct two Turing machines over the same fixed input alphabet. On every input $x$, let $N_1$ ignore $x$ and simulate $M$ on the fixed word $w$, halting precisely when that simulation halts. Let $N_2$ loop forever on every input.

If $M$ does not halt on $w$, both halting times are infinity for every $x$, so the pair belongs to $P$. If $M$ halts on $w$, the first halting time is finite while the second is infinity, so the pair does not belong to $P$. ThereforeThis is a many-one reduction from non-halting to the pointwise running-time domination problem. Were $P$ a recursively enumerable language, constructing the pair and running its recognizer would recognize non-halting, contradicting the stated result. Hence $P$ is not a recursively enumerable language. Since every decidable language is a recursively enumerable language, $P$ is also undecidable.

Give a nonhalting Turing machine computation halting time infinity. The formal language of pairs satisfying the displayed inequality is a co-recursively enumerable language but is not a recursively enumerable language. A violation has a finite witness: $N_1$ has halted on some input after $t$ moves while $N_2$ has not halted by that time. For the other direction, reduce non-halting by making $N_1$ simulate a fixed computation on every input and making $N_2$ run forever. The pair satisfies domination exactly when the fixed computation never halts. The convention $\infty \ge \infty$ is essential to this many-one reduction.