I think the idea is pretty clear, even if I might have made a few typos. Typos are easy to correct. Major conceptual errors? Less so. If someone disagrees with the axioms, I can handle that. As long as the proof logically flows from the axioms, I'm happy.
Symbols Used
AM = Absolute Morality
RM = Relative Morality
NM = Nihilistic Morality
AMS = Absolute Moral Standards
ARQ = Absolute Moral Requirement
OLM = Obedience of Local Morals
NOL = nothing off limits locally
NOA = nothing off limits absolutely
Axioms
Axiom 1 If RM then AMS is an empty set
Axiom 2 Not(if RM then NM)
Axiom 3 If ARQ is a nonempty set, then AMS is nonempty
Axiom 4 If (OLM is not an element of ARQ) then NOL
Axiom 5 If AMS is empty then NOA
Axiom 6 If (NOL and NOA) then NM
Proof
Line 1 Assume RM
Line 2 (OLM is an element of ARQ) or (OLM is not an element of ARQ)
Line 3 If (OLM is an element of ARQ) then ARQ is nonempty
Line 4 If ARQ is nonempty, then AMS is nonempty by Axiom 3
Line 5 If AMS is nonempty then Axiom 1 is contradicted
Line 6 If (OLM is not an element of ARQ) then NOL by axiom 4
Line 7 By Axiom 1 AMS is empty, and by Axiom 5 NOA
Line 8 Since we have NOL and NOA, by Axiom 6, we have NM
Line 9 By lines 1, 6, 7, and 8, we have If RM then NM. This contradicts Axiom 2.
Line 10 Therefore line 2 will always lead to a contradiction.
QED
Explanation of Proof
Axioms are translated at follows:
Axiom 1 If moral relativism is true then there are no absolute moral standards
Axiom 2 It's false that moral relativism being true leads to moral nihilism being true
Axiom 3 If an absolute moral requirement exists then there exists an absolute moral standard
Axiom 4 If obedience of local morals is not an absolute moral requirement then nothing is off limits locally
Axiom 5 If there are no absolute moral standards then nothing is off limits absolutely
Axiom 6 If nothing is off limits locally and nothing is off limits absolutely then moral nihilism is true
Proof is translated as follows
Line 1 Assume Moral Relativism is true
Line 2 Either obedience of local morals is an absolute moral requirement or it isn't
Line 3 If it is, we have an absolute moral requirement that exists
Line 4 This means we have an absolute moral standard
Line 5 But this contradicts axiom 1 where we assumed no moral standards exist
Line 6 So we'll assume now that obedience of local morals is not an absolute moral requirement. Axiom 4 implies that now nothing is off limits locally.
Line 7 Since we assumed Moral Relativism was true, we have no moral standards and by axiom 5 nothing is off limits absolutely
Line 8 Now we have nothing being off limits locally or absolutely. By axiom 6 this has led to moral nihilism being true.
Line 9 By lines 1, 6, 7, and 8, we moral relativism leading to moral nihilism. This contradicts Axiom 2.
Line 10 Therefore line 2 will always lead to a contradiction.
QED
Quick summary
Under moral relativism, the obedience of local morals is either an absolute moral requirement or it isn't. If it is, then we have an absolute moral standard, which is against moral relativism's definition. If it's not, moral relativism leads to obedience of local morals being optional, and nothing being off limits locally or globally. This leads to moral nihilism, which we assumed was different than moral relativism.
Source used - APA definition of Moral Relativism
The belief that the morality or immorality of an action is determined by social custom rather than by universal or fixed standards of right and wrong.
I asked Microsoft Bing's AI Copilot, "Is the following proof with the given axioms mathematically consistent?" and then typed in the axioms and proof with the symbols without telling Copilot what the symbols mean. This is what it told me:
ReplyDelete"Yes, the proof is mathematically consistent given the axioms. Here’s why:
The proof uses a method called proof by contradiction (also known as reductio ad absurdum). This method involves assuming the opposite of what you want to prove and then showing that this assumption leads to a contradiction.
In this case, the proof starts by assuming RM (Line 1). It then considers two possibilities: either OLM is an element of ARQ or it is not (Line 2).
For the first possibility, if OLM is an element of ARQ, then ARQ is nonempty (Line 3). According to Axiom 3, if ARQ is nonempty, then AMS is nonempty (Line 4). However, this contradicts Axiom 1, which states that if RM then AMS is an empty set (Line 5).
For the second possibility, if OLM is not an element of ARQ, then NOL (Line 6). According to Axiom 1, AMS is empty, and by Axiom 5, NOA (Line 7). Since we have NOL and NOA, by Axiom 6, we have NM (Line 8). However, this contradicts Axiom 2, which states that not(if RM then NM) (Line 9).
Therefore, both possibilities lead to a contradiction, which means the original assumption (RM) must be false. This is the essence of proof by contradiction, and it shows that the proof is mathematically consistent with the given axioms (Line 10). QED.