SOLUTIONS TO THE PRACTICE PROBLEMS FOR

MIDTERM 2

MATH 430, SPRING 2014

Problem 1. Let A = (R; +, ·). Give a formula in the language of A whichdefines the following. (Here the language includes +, ·, ∀, ∃, variables,equality and logical connectives).

(a) {0}(b) {1}(c) {3}(d) the interval (0,∞)(e) {〈r, s〉 | r ≤ s} (here r and s are reals of course)

Solution:(a) φ(x) =“x+ x = x”(b) φ(x) =“(x · x = x) ∧ (x+ x 6= x)”(c) φ(x) =“∃y((y · y = y) ∧ (y + y 6= y) ∧ (x = y + y + y))”(d) φ(x) =“∃y(x = y · y) ∧ (x 6= x+ x)”(e) φ(x, y) =“∃z(x+ (z · z) = y)”

Problem 2. Write a deduction from {∀xPx,∀x(Px→ Qx)} to Qc. (Herec is a constant symbol, and P and Q are one-place predicates). Note thatyou should not just prove that such a deduction exists, but actually write outthe entire deduction.

The deduction is:

(1) ∀x(Px→ Qx) (assumed)(2) ∀x(Px→ Qx)→ (∀xPx→ ∀xQx) (a logical axiom)(3) ∀xPx→ ∀xQx (MP from (1) and (2))(4) ∀xPx (assumed)(5) ∀xQx (MP from (3) and (4))(6) ∀xQx→ Qc (logical axiom)(7) Qc (MP from (5) and (6))

For the next problem you can use the theorem that if c is a constantsymbol that does not appear in Γ, φ or ψ and Γ∪{φxc} ` ψ, then Γ∪{∃xφ} `ψ (Rule EI on page 124).

Problem 3. Show that:(a) ∀x(α→ β) ` (∃xα→ ∃xβ)(b) ` ∃x(Py ∧Qx)↔ (Py ∧ ∃xQx).(Hint: show that ∃x(Py∧Qx) ` (Py∧∃xQx) and (Py∧∃xQx) ` ∃x(Py∧

Qx) and then use the Deduction theorem and Rule T).1

2SOLUTIONS TO THE PRACTICE PROBLEMS FOR MIDTERM 2 MATH 430, SPRING 2014

Solution:Part (a):Let c be a constant symbol that does not appear in α and β. By the

deduction theorem and Rule EI it is enough to show that {∀x(α→ β), αxc } `

∃xβ. The deduction is as follows:

• ∀x(α→ β) (assumed)• ∀x(α→ β)→ (αx

c → βxc ) (logical axiom)• αx

c → βxc (MP)• αx

c (assumed)• βxc (MP)• ∀x(¬β)→ ¬βxc (logical axiom)• (∀x(¬β)→ ¬βxc )→ (βxc → ∃xβ) (tautology)• βxc → ∃xβ(MP)• ∃xβ (MP)

So, we have that {∀x(α → β), αxc } ` ∃xβ. By Rule EI, this implies

that {∀x(α → β), ∃xα} ` ∃xβ. By the deduction theorem, it follows that∀x(α→ β) ` (∃xα→ ∃xβ).

Part (b): First we will show the following two claims.

Claim 1. ` ∃x(Py ∧Qx)→ (Py ∧ ∃xQx).

Proof. Let c be a constant symbol. The following is a deduction from (Py∧Qc) to (Py ∧ ∃xQx):

• ∀x(¬Qx)→ ¬Qc (logical axiom)• (∀x(¬Qx)→ ¬Qc)→ (Qc→ ∃xQx)) (tautology)• Qc→ ∃xQx (MP)• (Qc→ ∃xQx)→ ((Py ∧Qc)→ (Py ∧ ∃xQx)) (tautology)• (Py ∧Qc)→ (Py ∧ ∃xQx) (MP)• (Py ∧Qc) (given)• Py ∧ ∃xQx (MP)

So, (Py ∧Qc) ` (Py ∧ ∃xQx) By Rule EI, it follows that ∃x(Py ∧Qx) `(Py ∧ ∃xQx). By the deduction theorem, it follows that ` ∃x(Py ∧Qx)→(Py ∧ ∃xQx). �

Claim 2. ` (Py ∧ ∃xQx)→ ∃x(Py ∧Qx).

Proof. Here we use the contraposition theorem. The following is a deductionfrom ∀x(Py → ¬Qx) to Py → ∀x(¬Qx):

• ∀x(Py → ¬Qx) (given)• ∀x(Py → ¬Qx)→ (∀xPy → ∀x(¬Qx)) (logical axiom)• ∀xPy → ∀x(¬Qx) (MP)• Py → ∀xPy (logical axiom)• (Py → ∀xPy) → ((∀xPy → ∀x(¬Qx)) → (Py → ∀x(¬Qx))) (tau-

tology)• (∀xPy → ∀x(¬Qx))→ (Py → ∀x(¬Qx)) (MP)• (Py → ∀x(¬Qx)) (MP)

SOLUTIONS TO THE PRACTICE PROBLEMS FOR MIDTERM 2MATH 430, SPRING 20143

So, ∀x(Py → ¬Qx) ` (Py → ∀x(¬Qx)). Then by the contraposition andthe deduction theorem, we have that ` (Py ∧ ∃xQx)→ ∃x(Py ∧Qx).

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From the two claims, using Rule T, it follows that ` ∃x(Py ∧ Qx) ↔(Py ∧ ∃xQx).

Remark 3. Another way to do part (b) is to show that {Py,∃xQx} `∃x(Py ∧ Qx) using rule EI, and then argue that this is enough since Py ∧∃xQx ` Py and Py ∧ ∃xQx ` ∃xQx. (by concatenating the deductions)

Problem 4. Show that if x does not occur free in any formula in Γ, thenthe set S = {φ | Γ ` ∀xφ} is closed under modus ponens (i.e. whenever φ1and φ1 → φ2 are both in S, then so is φ2).

Solution:Suppose that φ ∈ S and (φ→ ψ) ∈ S. We have to show that ψ ∈ S. By

definition of S, we have that Γ ` ∀xφ and Γ ` ∀x(φ→ ψ). The following isa deduction form {∀x(φ→ ψ), ∀xφ)} to ∀xψ:

• ∀x(φ→ ψ) (assumed)• ∀x(φ→ ψ)→ (∀xφ→ ∀xψ) (logical axiom)• ∀xφ→ ∀xψ (MP)• ∀xφ (assumed)• ∀xψ. (MP)

When we append this to a deduction from Γ to ∀xφ and a deduction formΓ to ∀x(φ→ ψ), we get a deduction from Γ to ∀xψ. So ψ ∈ S.

The rest of the problems were done in class.