How to understand quantifier without predication ” ∀(λφ. (φ x m→ φ y))”?

I am reading about embedding/automation of modal logics in classical higher order logic ( and Goedels proof of God’s existence is prominent example here (as encoded for Isabelle/HOL).

This embedding has embedding for Leibniz equality for individuals:

abbreviation mLeibeq :: "μ ⇒ μ ⇒ σ" (infixr "mL=" 90) where "x mL= y ≡ ∀(λφ. (φ x m→ φ y))" 

and this type of euqality is used for the first axiom already:

A1a: "[∀(λΦ. P (λx. m¬ (Φ x)) m→ m¬ (P Φ))]" 

which can be written without lambdas as:

A1a: ∀φ[P(¬φ)↔¬P(φ)] 

My question is – how to understand the expression ∀(λφ. (φ x m→ φ y)), because usually we have ∀x.P(x)? I.e. universal quantifier expects the argument (x) and the predicate (P(x)), but this expression contains noone know what? is entire (λφ. (φ x m→ φ y)) and argument x or Predicate P(x)? What can be omitted here, what is the convention used here?

Discrete math – negate proposition using the quantifier negation

I’m asked to negate the following proposition using the quantifier negation rules. No negation operations are to appear before any of the quantifiers in the expression that is created. The issue is I’m not quite understanding what this means. All I’m given in my notes relating to negation quantifiers are the following formulas and their proofs:



I’m not quite sure how to take this information and apply it to a proposition or really, I don’t quite understand what my end goal of this question is supposed to be.

This is the proposition I’m given to work with. I’d appreciate if someone taught me step by step how to solve these types of questions. You can make up your own proposition if you’d like, but I’m really confused and would appreciate some sort of example. The results I found online seemed really complex and confusing.

∃𝑥 (𝐷(𝑥) → (𝐶(𝑥) ∨ F(x)))

Does quantifier elimination help here?

Suppose we have a quantified linear program

$ $ \exists z_1,\dots,z_{poly(n)}\in\mathbb R$ $ $ $ \exists u_1,\dots,u_n\in\mathcal P\cap\mathbb R^m$ $ $ $ \forall v_1,\dots,v_n\in\mathcal P\cap\mathbb R^m$ $ $ $ AX\leq b$ $ where $ \mathcal P$ is a bounded and convex polytope defined by $ poly(mn)$ inequalities then can quantifier elimination help reduce this to a single existential quantifier program with $ poly(mn)$ variables and inequalities?

Note every defining inequality is of degree $ 1$ and $ X$ is concatenation of all variables.

Real exponential field with restricted analytic functions: $\mathbb R_{an, exp, log}$ has quantifier elimination, but $\mathbb R_{an, exp}$ does not.

At a talk sometime ago a result was presented, which I believe originates from:

van den Dries, Lou; Miller, Chris, On the real exponential field with restricted analytic functions, Isr. J. Math. 85, No. 1-3, 19-56 (1994). ZBL0823.03017.

At some point it was mentioned that $ \mathbb R_{an,exp,log}$ admits quantifier elimination while $ \mathbb R_{an,exp}$ does not. Here $ \mathbb R_{an,exp}$ is the theory of the (ordered) real exponential field with function symbols for all restricted analytic functions. Then of course $ \mathbb R_{an,exp,log}$ is just adding a function symbol for logarithms.

Someone in the audience remarked that $ log(x)$ (or more precisely, its graph) is quantifier-free definable by $ x = exp(y)$ . Then a fairly simple formula was presented to show why you really need $ log$ as a function symbol for quantifier elimination, and there is my question: I just cannot remember or reconstruct that formula. So what would be a simple example of some formula in this setting that is not equivalent to a quantifier-free formula in $ \mathbb R_{an,exp}$ ?

I am probably missing something obvious here, but now it’s haunting me.

Does quantifier elimination help here in solving quantified convex program?

I have a quantified convex program of the form that I need to solve

$ $ \exists(x_{1,1},\dots,x_{1,n})\in\mathbb R^n\quad\forall(x_{2,1},\dots,x_{2,n}\in\mathbb R^n$ $ $ $ \vdots$ $ $ $ \exists(x_{2t-1,1},\dots,x_{2t-1,n})\in\mathbb R^n\quad\forall(x_{2t,1},\dots,x_{2t,n}\in\mathbb R^n$ $ $ $ \phi_1(x_{1,1},\dots,x_{2t,n})\leq a_1\wedge\dots\wedge\phi_r(x_{1,1},\dots,x_{2t,n})\leq a_r$ $ where $ \phi_1,\dots,\phi_r$ are convex polynomials of degree $ d$ with all polynomials and $ a_1,\dots,a_r$ in $ \mathbb Z$ coefficients and having at most $ m$ bits each.

I feel quantifier elimination over reals will help here. However how to go about it is unclear.

  1. Does quantifier elimination tell anything about the time complexity of the program and if so what is the complexity?

  2. Is it possible to use quantifier elimination to reduce the number of quantifications and if so how does the parameters change?

How to remove a universal quantifier in Lean theorem prover

I am working with two binary relations: g_o and pw_o, and I’ve defined pw_o below:

constants {A : Type} (g_o : A → A → Prop)

def pw_o (x y : A) : Prop := ∀ w : A, (g_o w x → g_o w y) ∧ (g_o y w → g_o x w)

I need to prove the following:

theorem prelim: ∀ x y z : A, g_o x y ∧ pw_o y z → g_o x z :=

I start with these tactics:

begin     intros,     cases a with h1 h2,  end 

And I have this:

x y z : A, h1 : g_o x y, h2 : pw_o y z ⊢ g_o x z 

Since pw_o is defined with a universal quantifier, I’d like to substitute w with x, then I would have (g_o x y → g_o x z) ∧ (g_o z x → g_o y x). After isolating the first conjunct with the “cases” tactic, I can use modus ponens on that first conjunct and h1. How can I tell Lean to replace w in the definition of pw_o with x and replace x and y in the definition of pw_o with y and z, respectively?