1
No one has power over p is a box
Read a world that w sees as a way things could still go, given what anyone can do at w. Then □p says that p holds in every such way: no one can make it false. The kripke lab computes this box exactly as before.
Lesson · Identity, Modality and Freedom · Freedom
The Consequence Argument
If determinism is true, our acts are the consequences of the laws and the remote past, and no one has power over those. Reading “no one has power over” as a box, the step that carries unavoidability across a consequence is the K axiom, valid on every frame the reader can draw. The kripke lab checks it and then lets you falsify each premise in a model.
N p: p, and no one has power over p
box reading: p at every world left open
□p ∧ □(p → q) → □q valid on every frame
premise one: the past and laws are fixed
premise two: determinism makes the act follow
Powerlessness passes across a consequence
Core ideas
What this lesson establishes
The previous lesson left the incompatibilist's conditional unsupported. This one gives it its best support, an argument whose logic can be checked, and shows where the checking stops.
2
The transfer step is K
If no one has power over p, and none over the fact that p leads to q, then none over q. That is □p ∧ □(p → q) → □q, the axiom called K, and the lab finds it valid on every frame.
3
What is left is the premises and the reading
A valid argument leaves you its premises to examine. Here there are two, and a third thing to examine: whether “has no power over” is the sort of operator the box computes.
The material
A valid transfer, and where to push
The argument as its defender states it, the check that its logic is K, and the three places a reader can push.
Let h be the state of the world long before anyone was born, l the laws of nature, and a some act of mine tomorrow. Suppose determinism is true: the remote past and the laws entail the act. The argument then runs in three lines. Nobody has any power over the remote past and the laws, since they were fixed before anyone could act. Nobody has any power over the fact that the past and the laws entail the act, since that is a matter of what the laws are. So nobody has any power over the act. If I have no power over my acts, I do not act freely.
Write N for “and no one has, or ever had, any power over whether”. The premises are N(h ∧ l) and N((h ∧ l) → a), and the conclusion is N a. The step from the premises to the conclusion is an instance of one rule: from N p and N(p → q), conclude N q. Peter van Inwagen called it the transfer of powerlessness. Since N p says p among other things, a frame that models N has each world seeing itself, and the presets that evaluate a premise give w1 that arc; the transfer step, as the next paragraph shows, needs no arc at all.
The lab's box is N when the arcs are read as described: world w sees a world when that world is still open, given what anyone can do. The transfer rule is then the formula in the key, and the lab can test it. Take any frame at all; the reason it holds is short. If p holds at every world seen from w, and p → q does too, then at each seen world p is true and p → q is true, so q is true, and q holds at every seen world. Nothing was assumed about the arcs.
Theorem · The transfer rule is valid on every frame
For every frame, every valuation and every world, □p ∧ □(p → q) → □q is true. The lab's first preset evaluates it at every world of a three-world frame and finds it true at all of them, and changing the arcs or the valuation does not change that.
So the argument is valid, given the reading of its box. A reader who wants to resist the conclusion has three choices, and the lab can model each. The first is to deny that the past and laws are beyond everyone's power. On David Lewis's compatibilism, had I done otherwise, a small miracle would have broken the laws shortly before. I cannot cause that miracle, but I can act in a way such that, were I to act so, it would have occurred, and that is a power over the laws of the relevant kind. In the second preset, w1 sees a world where the laws fail, and □(h ∧ l) is false at w1. This is the compatibilist's move.
The second choice is to deny that the past and the laws entail the act: to deny determinism. In the third preset the seen worlds all keep the past and the laws, and in one of them the act does not occur. Then □((h ∧ l) → a) is false at w1. This is the libertarian's move. The fourth preset shows both premises true and the conclusion true with them, as the validity of the step promises.
The third place to push
The last choice is not about either premise. It is to say that “no one has power over” does not behave as the box does. Every box satisfies agglomeration: from □p and □q comes □(p ∧ q), since a world where each holds at every seen world is one where the conjunction does. The operator N has been argued not to. A coin I could toss lies untossed. I have no power over whether it fails to land heads, since even tossing it cannot make it land heads; I have none over whether it fails to land tails, for the same reason; but I have power over whether it fails to do both, since tossing it would make one of them happen. If that is right, N fails agglomeration, no box does, and the transfer step is not an instance of K, because the operator is not a box; the lab's verdict does not apply. The argument's defender has replied by redefining N so that it is a box after all, as truth throughout every region of possibility anyone can reach, and then the two premises have to be argued for under that definition.
This answers the objection that the logic can be denied. K cannot be denied for a box. What can be denied is that the operator in the argument is a box, and that is a claim about the premises' meaning, not about the logic of necessity. The lab shows the logic is secure and is silent on whether the argument has the logic's shape.
Interactive
Possible worlds
Modal formulas at every world, and the axioms the frame validates
Method
Testing an argument that transfers a modal property
Five moves for any argument of the form: this is beyond control, that follows from it, so that is too.
1
Name the operator and what the arcs mean
Say what the box stands for here and what it is for one world to see another. The reading is part of the argument, and the lab cannot supply it.
2
Put the argument in the form of the transfer rule
Write the first premise as □p, the second as □(p → q) and the conclusion as □q. If the argument does not have that shape, K does not apply.
3
Test the rule on a frame of your own
Build a frame, type the formula and read the worlds tile. It is valid on every frame, so any frame will do; a failure would show a mistake in the typing.
4
Falsify each premise in a model
Draw the smallest model in which the first premise is false and the smallest in which the second is. Each model is a view of the world, and each has a holder.
5
Ask whether the operator is a box
If the reader's has no power over fails to combine as a box does, the argument has lost its logic. Say which instance of the rule fails, with a case.
Powerlessness across a consequence
Score 0 / 0Choose the answer you can justify.
Both premises and the conclusion, at w1
Two worlds. World w1 sees itself and w2. The past and laws hold in both, and the act occurs in both.
h and l at w1 and at w2: true
so box of (h and l) at w1: true
(h and l) implies a at w1 and at w2: true
so box of ((h and l) implies a) at w1: true
a at w1 and at w2: true, so box of a at w1: true
the act is beyond power, as the argument said
now add a seen world w3 without the act
box of ((h and l) implies a) at w1 fails
The conclusion followed from the premises, as the rule says it must. The last two lines show a model in which a premise fails, and in it the conclusion fails too. That is a libertarian's picture, and the lab does not say it is the right one.
Common mistakes
1
Believing the consequence argument can be escaped by denying its logicThe transfer rule is K, and the lab finds it true at every world of every frame, however the arcs and atoms are changed. A reader who wants to resist has to deny a premise or deny that the operator is a box, and these are claims about the world and about the argument's reading, not about the logic of the box.
2
Taking the validity of the transfer step to show that the premises are trueThe lab's fourth preset has both premises true because they were typed in as true. The second and third make one false. In each model the step is valid, and the models differ in what the world is like. Whether the past and laws are beyond everyone's power, and whether the laws entail the act, are what the dispute is about.
3
Drawing the arcs of N to every world there isIf w1 sees every logically possible world, it sees worlds with another past and other laws, □(h ∧ l) is false at w1, and the argument is lost at its first premise before anyone has argued against it. The arcs of N are narrower: w1 sees a world when someone at w1 could still bring it about. Under that reading the first premise says that no one can bring about a different past or different laws, which is the claim Lewis's compatibilist contests, and the lab draws whichever arcs you give it.
Check your understanding
✓
Finish when you can show that the transfer step is valid on every frame, and name the premise or the reading each response rejects.
You should be able to write the argument as a box over two premises and a conclusion, test the transfer rule on a frame you built, draw a model in which each premise fails, and state which position the model represents. A response that says only that the argument is “question-begging” has not located the premise.
Note: Worlds w2 and w3 see nothing in the models of the second and third presets, so a box is vacuously true there, and the verdict to read is the one at w1. The lab computes with whichever arcs you give it. It cannot say that the past is fixed, that the laws fix the act, or that the arcs you drew are the ways things are still open to anyone; the argument turns on those.