# The Metaphysical Puzzle Of The Ship Of Theseus: Persistence In Time - King's College London - Essay

1289 words - 6 pages

What is the puzzle of the ship of Theseus? Outline and evaluate one potential solution.
The aim of this paper is to give a possible solution for solving the puzzle of the ship of Theseus. I will first outline what the puzzle is about and its main metaphysical implications, and I will then proceed by taking into consideration one of the possible solutions, namely perdurantism and four dimensionalism. I will explain what being a four-dimensional object implies, and give an interpretation of why this might be a possible solution for the puzzle.
The puzzle of the ship of Theseus is a philosophical puzzle that has to do with problems associated with numerical identity and the concept of transitivity. The puzzle goes as follows. The ship that once belonged to Theseus was left in the Athens’ harbour for many years. In order to prevent it from deteriorating, once a year one plank was substituted with a new plank. It is now necessary to state that, for the purpose of this paper, the elements that constitute the ship will be reduced to the planks only, as this does not affect the consequences of the puzzle but it highly simplifies its understanding. Therefore, after 1000 years, the ship of Theseus, that will be referred to as ship A, or original ship, will have undergone a complete change of its components. I shall call this renewed ship ship B. The planks that were taken away from ship A were not dismantled but were preserved and reassembled to reconstruct a new ship, ship C, that was conserved in the local museum of Athens for visitors to see.
Therefore, on one hand, we have that A = B, because it is easily graspable that a ship can undergo a process of repair whilst still being the same ship. On the other hand, we hold that A = C, because a ship (C) made up of the same components as the original ship (A) is also the same ship as the original (A). However, and this is the core of the puzzle, there can’t be two ships of Theseus, namely B and C. As stated above, the problem is twofold. Firstly, it is a problem concerning numerical identity. Numerical identity holds that things are numerically identical if they are one and the same thing; but, according to this definition, both B and C are one and the same with A. Secondly, it is a problem of transitivity. If A = B, and A = C, then, by property of transitivity, B = C; however, we know that ship B and ship C are two different ships, being one in the harbour and the other in the museum.
One way of solving the puzzle is resorting to perdurantism and four dimensionalism.[footnoteRef:1] Before delineating how these concepts can, in actual fact, come as a helpful means in understanding the puzzle, I will give a definition of both. Four dimensionalism holds that objects are constituted not only of spatial parts but also of temporal parts, or stages. In this sense, they are four dimensional, being spread out not only in the three dimensions of space – height, width, depth – but also in time. Following from four dimensionalism, perdurantism holds that “an object persists through time in virtue of possessing successive temporal parts, so that [...] only part of a persisting object, its current temporal part, is present at any one time during its existence.” (Lowe 2002). Thus, objects’ persistence over time is possible because they have different temporal parts at distinct times. In a sense, it is as if an object is never fully revealed at different stages, but each time just a part of the object is revealed. Hence, for perdurantists, an object at t0 is the same at t1, but the part of the object shown at t0 is different from the part of the object shown at t1. [1: Ney, Alyssa, Metaphysics: an Introduction, p. 173-177. ]
This is in contrast with another view, that of endurantism, that is based on three dimensionality: objects are described in having only a spatial extension, and therefore they “persist over time by being wholly present at each time at which they exist”.[footnoteRef:2] [2: Ney, Alyssa, Metaphysics: an Introduction, cit. p. 174. ]
I will now examine how perdurantism and four dimensionalism link back to the puzzle. The reason for the incongruence before, namely that a) A was numerically identical both to B and C and b) B and C were different even though should have been the same in the name of transitivity, is that we held an endurantist, three dimensional view. This means that we thought A persisted over time by being a later object but always identical to it; so ship B was a later ship A, because it was identical to it, and also C was a later A because it was identical too.
However, if we change our perspective and analyse the puzzle through the lenses of perdurantism and four dimensionalism, then we have that A’s persistence over time is linked to its different temporal parts at different times and is not wholly present at each time; therefore, A needs not to be strictly identical to B or C.
In order to fully understand the puzzle, we need to clarify another step. Following Ney’s argument,[footnoteRef:3] most of perdurantists and four dimensionalists are also mereological universalists, meaning that they believe that “any non-overlapping material objects whatsoever compose some further object.”.[footnoteRef:4] Hence, we have two temporally extended objects: 1) one that includes A and B, and 2) another that includes A and C in their parts. Since both contain A’s parts, there is not a best candidate to be A. While this constituted a problem before, because both B and C were identical to A, here there is no such problem. Indeed, we don’t have to choose between B and C in the first place, because A’s persistence over time is not related to its strict numerical identity. [3: Ney, Alyssa, Metaphysics: an Introduction, p. 175.] [4: Ney, Alyssa, Metaphysics: an Introduction, cit. p. 175-176. ]
An objection might arise at this point. If we are to say that both B and C share a part of A, then we would need to accept that B and C are present in the same location at t0, meaning that there are two ships to start with in the same location at the same time. However, this, once again, does not constitute a problem. Indeed, perdurantists do not hold that there are two ships wholly present at the same location at the same time, but simply that a temporal part of each ship is present at the same location at the same time. This constitutes a case of temporal overlap, which is not different than a spatial overlap, that is easily understood in many other cases, e.g. roads overlapping when sharing a common part while still carrying their own name.
In conclusion, the puzzle of the ship of Theseus can be solved by resorting to perdurantism and four dimensionalism since this would solve both a) the problem of numerical identity and b) that of transitivity. Indeed, a) we wouldn’t have to choose between B and C in the first place, since for an object to persist over time it does not need to be strictly identical, and b) we can allow both B and C to be the same as A but different in regard to each other by saying that they only share a common temporal part (A) when they overlap.
(word count: 1254)
Bibliography:
Lowe, J. 2002. A Survey of Metaphysics. New York: Oxford University Press.
Ney, Alyssa. 2014. Metaphysics: an Introduction. New York: Routledge.
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