10 / 18 · Think in objects and relations
All, some, and finite worlds
Evaluate quantified statements in a specified, nonempty domain.
Builds on When the premises cannot all hold
Go to practice ↓A question to keep in mind
'All the reviewed papers are public.' Which papers count, and must any be reviewed?
A predicate describes an object: R(x) may mean 'x is red'. A domain specifies which objects x can name. In our small world the domain is exactly a, b, c; R means red and C means circle. Neither predicate says anything about objects outside this world.
∀x C(x) says every object in the domain is a circle. ∃x C(x) says at least one is. Existence is established by a witness: name an object for which the predicate is true. A universal claim fails with just one counterexample object.
'All red objects are circles' is ∀x(R(x) → C(x)). 'Some red object is a circle' is ∃x(R(x) ∧ C(x)). The universal restricts by a conditional; the existential needs an object satisfying both properties. An object that is not red does not violate the universal.
Work through an example
- a is a red circle; b is a gray square; c is a gray circle. Every red object is a circle, but not every object is a circle.
- a witnesses ∃x(R(x) ∧ C(x)). b refutes ∀x C(x). These are different roles played by named objects.
- Checking this world proves what is true here, not what is true in all possible domains.
Your turn
0 / 3Build a world with a red circle, but not every object a circle.
Read solution · does not award completion
One possible world:
- a: red=T, circle=T
- b: red=F, circle=F
- c: red=F, circle=F
One red circle supplies a witness. A different non-circle defeats the universal. The same object cannot be both a circle and a non-circle here.
Which means 'Some red object is a circle'?
Read solution · does not award completion
∃x(R(x) ∧ C(x))
For an existential conjunction the two properties must meet in a single object.
A formula holds in every world you tried with three objects. What is established?
Read solution · does not award completion
It holds in those tested worlds; general validity needs more.
A finite countermodel can disprove validity. Finite successes do not in general prove first-order validity.
Bring it back to your own work
Translate a claim beginning with 'all' or 'some' from a paper. State the domain and what would count as a witness or counterexample.