Episode 10. Blind quantifiers. Unistructurality The blind universal and existential quantifiers DeMorgan’s laws for blind quantifiers The hierarchy of quantifiers How blind quantification affects game trees. 0. 10.1. Unistructurality.
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A game A is said to be unistructural iff, for any valuations e1 and e2, we have
Intuitively, unistructural games are games all instances of which have the same
structure. All of the examples of games that we have seen so far are unistructural, and
so are all naturally emerging games. Also, all of our game operations preserve the
unistructural property of games. Hence, if you wish, it is safe to assume that “game”
always means “unistructural game”. For the sake of generality, however, we do not
officially impose this unnecessary restriction on games.
Our definition (next slide) of the blind quantifiersx (universal) and x (existential),
however, assumes that the game A(x) to which they are applied satisfies the weaker
condition of x-unistructurality.
We say that a game A(x) is x-unistructural iff, for any valuation e and any two
constants c and d, Lre[A(c)]=Lre[A(d)]. Intuitively, this means that the structure of
any given instance of A does not depend on the value of x (but may depend on the
values of some other variables).
Of course, every unistructural game is also x-unistructural for any variable x.
Definition 10.1. Let A(x)be an x-unistructural game.
(a) The gamexA(x)is defined by:
(b) The gamexA(x) is defined by:
Intuitively, playing xA(x) or xA(x) means just playing A(x) “blindly”, without
knowing the value of x. In xA(x), Machine wins iff the play it generates is successful
for every possible value of x, while in xA(x) being successful for just one value is
When applied to elementary games, the blind quantifiers, just like the parallel
quantifiers, act exactly as the quantifiers of classical logic.
Unlike xA(x) which is a game on infinitely many boards, both
⊓xA(x)and xA(x) are one-board games. Yet, they are very different
from each other.
To see this difference, compare the problems
2 (Environment selects a
number m, to which Machine
replies by 0 or 1)
1 (only Machine has a
move, 0 or 1)
Of course. Evenness
is a decidable problem.
No. Machine cannot
select a ⊔-disjunct
that would be true
for all values of x.
x(Even(x)⊔Odd(x) ⊓y(Even(x+y)⊔Odd(x+y)))A computable -problem
This problem is computable. The idea of a winning strategy is that, for any given y,
in order to tell whether x+y is even or odd, it is not really necessary to know the value
of x. Rather, just knowing whether x is even or odd is sufficient. And such knowledge
can be obtained from the antecedent.
In other words, for any known y and unknown x, the problem of telling whether
x+y is even or odd is reducible to the problem of telling whether x is even or odd.
Specifically, of both x and y are even or both are odd, then x+y is even; otherwise
x+y is odd.
Below is the evolution sequence induced by a run where Machine used a winning
The play hits ⊤, and thus Machine wins.