NameTheTrait 2.0
This article is intended to be an improvement on the original (logically invalid) NameTheTrait argument. It is possible to correct the NTT argument and preserve its persuasive force by adding a premise that rejects double standards and by changing the first premise to require human moral value (or some other moral consideration) to be based on a trait. This makes NTT valid, and allows it to be presented in the same way as it was intended.
Contents
NTT 2.0 In English
In the following x has R means; we are moral required to not consume the products of x for food, in order to prevent harm to x.
(This could be substituted for moral value or the right to equal consideration of interests etc.)
(P1) There exists a trait, such that, if an individual is a human then the individual has R if and only if they possess the trait
- (humans who possess a certain trait have R (this may or may not include all humans))
(P2) If humans have R if and only if they possess a certain trait, then all beings have R if an only if they possess the certain trait.
- (rejection of double standards)
(P3) If an individual is a sentient nonhuman animal, then there is no trait absent in the individual, which if absent in a human, would cause the human to not have R.
or equivalently
(P3) If an individual is a sentient nonhuman animal, then the trait that grants humans R is present in the individual.
- (sentient nonhuman animals possess this certain trait)
Therefore
(C) Sentient nonhuman animals have R
- (we are morally required to go vegan)
Defense of Premises
P1 - All that is necessary here is to convince the opponent that humans who have R, have R in virtue of some trait that they possess, it is not necessary to specify which trait. But examples may include example sentience, capacity for sentience, moral agency etc. Furthermore all humans may or may not possess this trait.
P2 - Most people would reject moral double standards, it is unlikely that one would face much resistance on this premise.
P3 - This premise is the one which is likely to be contended. However if one can convince an opponent that the sentience, capacity for sentience or similar, is what grants humans R, then they must accept the premise, as sentient animals possess these traits. A strategy to get people to reach this trait, is to consider traits commonly used as justifications for denying animals moral status or in our case R, such as (low) intelligence, moral agency etc. Now if someone accepts P2 (rejects double standards), then they would also have to conclude that humans who lack moral agency, (low) intelligence etc. do not have moral status, or in our case R, which they are unlikely to actually accept. This is the essence of the argument, and this strategy has been claimed to be a effective by activists.
Informal Presentation
In the case of making informal arguments, which rely on implicit agreements and assumptions between opponents, such as when presenting arguments to friends, family or street activism etc. it is not necessary to present the argument in full detail. One way to present the argument, which is used by activists, is the following:
- vegan : can you name a trait present or absent in animals that justifies treating them the way we do
- opponent : moral agency, (low) intelligence etc.
- vegan : what about humans who lack moral agency, have (low) intelligence etc. such as the mentally disabled and infants, would it be justified to treat them in this way?
- opponent : wait.. no..
- vegan : You see, I think it's the capacity to experience well-being, happiness, and suffering (sentience) that gives a being moral status, and means we shouldn't harm them for our taste pleasure. Would you agree with that?
- opponent : Yes, I suppose I would. Perhaps going vegan would be the right thing to do
In First Order Logic
Definitions
- H(x) means 'x is a human'
- SNA(x) means 'x is a sentient non-human animal'
- R(x) means 'we are morally required to seek to exclude—as far as is possible and practicable (AFAPP)—all forms of exploitation of, and : cruelty to, x'
- T(x) means 'x is a trait'
- P(x,y) means 'x has y'
In First Order Logic
- (P1) ∃t ( Tt ∧ ∀x ( Hx ⇒ ( Rx ⇔ Pxt ) ) )
- (P2) ∀t ( Tt ∧ ( ∀x ( Hx ⇒ ( Rx ⇔ Pxt ) ) ⇒ ∀x ( Rx ⇔ Pxt ) ) )
- (P3) ∀x( SNAx ⇒ ¬∃t ( Tt ∧ ¬Pxt ∧ ∀y ( Hy ⇒ ( ¬Pyt ⇒ ¬Ry ) ) ) )
or equivalently
- (P3) ∀x( SNAx ⇒ ∀t ( Tt ∧ ∀y ( Hy ⇒ ( Ry ⇒ Pyt ) ) ⇒ Pxt ) )
- Therefore (C) ∀x ( SNAx ⇒ Rx )
Direct Translation
- (P1) there exists t, such that t is a trait, and for all x, if x is a human, then x has R, if and only if x has t
- (P2) for all t, t is a trait, and if for all x, if x is a human, then x has R if and only if x has t, then for all x, x has R if and only if x has t
- (P3) for all x, if x is a sentient nonhuman animal, then there does not exist t, such that, t is a trait and, x lacks t, and for all y, if y is a human, then if y lacks t, then y does not have R
or equivalently
- (P3) for all x, if x is a sentient non-human animal, then for all t, if t is a trait, and, for all y, if y is a sentient human, then, if y has R then y has t, then x has t
- Therefore (C) for all x, if x is a sentient nonhuman animal then x has R
Proof of Validity
Logical Proof Generator
We can show this is valid argument by using a logical proof generator, to prove the formula
- P1 ∧ P2 ∧ P3 ⇒ C
with the input
- (P1) \existst ( Tt \land \forallx (Hx \to ( Rx \leftrightarrow Pxt ) ) )
- (P2) \forallt (Tt \land ( \forallx (Hx \to ( Rx \leftrightarrow Pxt ) ) \to \forallx ( Rx \leftrightarrow Pxt ) ) )
- (P3) \forallx ( Ax \to \neg \existst ( Tt \land \negPxt \land( \forally (Hy \to ( \negPyt \to \negRy ) ) ) ) )
or equivalently
- (P3) \forallx( Ax \to \forallt ( Tt \land \forally ( Hy \to ( Ry \to Pyt ) ) \to Pxt ) )
- (C) \forallx ( Ax \to Rx )
Or all together (P1 ∧ P2 ∧ P3 ⇒ C)
- \existst ( Tt \land \forallx (Hx \to (Rx \leftrightarrow Pxt ) ) ) \land \forallt ( Tt \land ( \forallx (Hx \to ( Rx \leftrightarrow Pxt)) \to \forallx (Rx \leftrightarrow Pxt))) \land \forallx (Ax \to \neg \existst (Tt \land \negPxt \land (\forally (Hy \to ( \negPyt \to \negRy))))) \to \forallx ( Ax \to Rx )
which yields valid
Note that without the modification of premise 1 to require human moral value to be based on trait, and the addition of the premise to forbid double standards, the argument would not be even close to valid, which is the case for the original NTT formulation.
Natural Deduction
First we will prove the following to make our proof simpler:
∀x( SNAx ⇒ ¬∃t ( Tt ∧ ¬Pxt ∧ ∀y ( Hy ⇒ ( ¬Pyt ⇒ ¬Ry ) ) ) ) ⇔ ∀x( SNAx ⇒ ∀t ( Tt ∧ ∀y ( Hy ⇒ ( Ry ⇒ Pyt ) ) ⇒ Pxt ) )
Starting with the left-hand-side (LHS)
LHS | ⇔ ∀x( SNAx ⇒ ¬∃t ( Tt ∧ ¬Pxt ∧ ∀y ( Hy ⇒ ( ¬Pyt ⇒ ¬Ry ) ) ) ) | |
⇔ ∀x( SNAx ⇒ ∀t ¬( Tt ∧ ¬Pxt ∧ ∀y ( Hy ⇒ ( ¬Pyt ⇒ ¬Ry ) ) ) ) | (¬∃x Px) ⇔ (∀x ¬Px) | |
⇔ ∀x( SNAx ⇒ ∀t( Tt ∧ ¬Pxt ⇒ ¬∀y ( Hy ⇒ ( ¬Pyt ⇒ ¬Ry ) ) ) ) | ¬(p∧q) ⇔ (p⇒¬q) | |
⇔ ∀x( SNAx ⇒ ∀t( Tt ∧ ¬Pxt ⇒ ¬∀y ( Hy ⇒ ( Ry ⇒ Pyt ) ) ) ) | (p⇒q) ⇔ (¬q⇒¬p) | |
⇔ ∀x( SNAx ⇒ ∀t( Tt ∧ ∀y ( Hy ⇒ ( Ry ⇒ Pyt ) ) ⇒ Pxt) ) | (p⇒q) ⇔ (¬q⇒¬p) | |
⇔ RHS |
Hence we have shown the equivalency.
Now we can prove the validity of the argument using natural deduction.
1 | ∃t ( Tt ∧ ∀x ( Hx ⇒ ( Rx ⇔ Pxt ) ) ) | assumption (P1) | |
2 | ∀t ( Tt ∧ ( ∀x ( Hx ⇒ ( Rx ⇔ Pxt ) ) ⇒ ∀x ( Rx ⇔ Pxt ) ) ) | assumption (P2) | |
3 | ∀x( SNAx ⇒ ∀t ( Tt ∧ ∀y ( Hy ⇒ ( Ry ⇒ Pyt ) ) ⇒ Pxt ) ) | assumption (P3) | |
4 | Ts ∧ ∀x ( Hx ⇒ ( Rx ⇔ Pxs ) ) | existential elimination | 1 |
5 | Ts ∧ ( Ha ⇒ ( Ra ⇔ Pas ) ) | universal elimination | 4 |
6 | Ts ∧ ( ( Ha ⇒ ( Ra ⇔ Pas ) ) ⇒ ( Rb ⇔ Pbs ) ) | universal elimination | 2 |
7 | SNAb ⇒ ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ⇒ Pbs ) | universal elimination | 3 |
8 | ¬ ( SNAb ∧ ¬ ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ⇒ Pbs ) ) | ( p ⇒ q ) ⇔ ¬( p ∧ ¬q ) | 7 |
9 | ¬ ( SNAb ∧ ¬ ¬( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ∧ ¬ Pbs ) ) | (p ⇒ q) ⇔ ¬(p ∧ ¬q) | 8 |
10 | ¬ ( SNAb ∧ ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ∧ ¬ Pbs ) ) | negation elimination | 9 |
11 | ¬ ( SNAb ∧ ¬ Pbs ∧ ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ) ) | ∧ commutivity | 10 |
12 | SNAb ∧ ¬ Pbs ⇒ ¬ ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ) | ¬( p ∧ q) = (p ⇒ ¬q) | 11 |
13 | ¬ (SNAb ⇒ Pbs) ⇒ ¬ ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ) | ¬(p ⇒ q) ⇔ (p ∧ ¬q) | 12 |
14 | ( Ts ∧ ( Ha ⇒ ( Ra ⇒ Pas ) ) ) ⇒ (SNAb ⇒ Pbs) | (¬p ⇒ ¬q) ⇔ (q ⇒ p) | 13 |
15 | Ts ∧ ¬( Ha ∧ ¬ ( Ra ⇔ Pas ) ) | (p⇒q) ⇔ ¬(p∧¬q) | 5 |
16 | Ts ∧ ¬( Ha ∧ ¬ ( (Ra ⇒ Pas) ∧ (Pas ⇒ Ra) ) ) | biconditional elimination | 15 |
17 | Ts ∧ ¬ ( Ha ∧ (¬ (Ra ⇒ Pas) ∨ ¬(Pas ⇒ Ra) ) ) | ¬(p ∧ q) = ¬p ∨ ¬q | 16 |
18 | Ts ∧ ¬ ( (Ha ∧ ¬ (Ra ⇒ Pas) ) ∨ (Ha ∧ ¬( Pas ⇒ Ra ) ) ) | p ∧ (q ∨ r ) = (p ∧ q) ∨ (p ∧ r) | 17 |
19 | Ts ∧ ¬ (Ha ∧ ¬ (Ra ⇒ Pas) ) ∧ ¬ (Ha ∧ ¬( Pas ⇒ Ra ) ) | ¬ (p ∨ q) = (¬p ∧ ¬q) | 18 |
20 | Ts ∧ (Ha ⇒ (Ra ⇒ Pas)) ∧ (Ha ⇒ (Pas ⇒ Ra) ) | ( p ⇒ q ) ⇔ ¬ ( p ∧ ¬q ) | 19 |
21 | Ts ∧ (Ha ⇒ (Ra ⇒ Pas)) | ∧ elimination | 20 |
22 | SNAb ⇒ Pbs | Modus Ponens | 14, 21 |
23 | Rb ⇔ Pbs | Modus Ponens | 6, 5 |
24 | (Rb ⇒ Pbs) ∧ (Pbs ⇒ Rb) | biconditional elimination | 23 |
25 | Pbs ⇒ Rb | ∧ elimination | 24 |
26 | SNAb ⇒ Rb | transitivity | 22, 25 |
27 | ∀x (SNAx ⇒ Rx) | universal introduction (C) | 26 |