I'm learning about logic by going through the excellent Stanford Introduction to Logic, as part of writing a maths tutorial. Here I'm working through Exercise 3.4 - Propositional Entailment. Let me know if you disagree with any of my answers:

Let Γ and Δ be sets of sentences in Propositional Logic, and let φ and ψ be individual sentences in Propositional Logic. State whether each of the following statements is true or false.

a. If Γ ⊨ φ and Δ ⊨ φ, then Γ ∩ Δ ⊨ φ.

I'll show it's false by providing a counterexample. If P is an atomic formula:

Γ = {(P and not P), P}
Δ = {(φ and not φ), P}

Since Γ and Δ contain an unsatisfiable formula they entail everything, so Γ ⊨ φ and Δ ⊨ φ are both true. Also:

Γ ∩ Δ = {P}

The formula (P implies φ) is not valid, and so it's false that Γ ∩ Δ ⊨ φ.

b. If Γ ⊨ φ and Δ ⊨ φ, then Γ ∪ Δ ⊨ φ.

If Γ = {P1, P2, P3, ...} then:

(P1 and P2 and P3 and ...) implies φ

is valid. If Δ = {Q1, Q2, Q3, ...}

(P1 and P2 and P3 and ...) and (Q1 and Q2 and Q3 and ...) implies φ

must also be valid since for any interpretation, adding more 'and' terms to a formula can never make it true if it was false. If adding the Q terms only introduces false interpretations on the left hand side of the 'implies' then the whole formula will remain valid. So:

Γ ∩ Δ ⊨ φ

is true.

c. If Γ ⊨ φ and Δ ⊭ φ, then Γ ∪ Δ ⊨ φ.

Using the same reasoning as above, if:

Γ = {P1, P2, P3, ...}

then:

(P1 and P2 and P3 and ...) implies φ

is valid. If:

Δ = {Q1, Q2, Q3, ...}

then:

(P1 and P2 and P3 and ...) and (Q1 and Q2 and Q3 and ...) implies φ

must also be valid since for any interpretation, adding more 'and' terms to a formula can never make it true if it was false. If adding the Q terms only introduces false interpretations on the left hand side of the 'implies' then the whole formula will remain valid. So:

Γ ∪ Δ ⊨ φ

is true.

d. If Γ ⊭ ψ, then Γ ⊨ ¬ψ.

If:

Γ = {P1, P2, P3, ...}

then if Γ ⊭ ψ then:

(P1 and P2 and P3 and ...) implies ψ

is not valid.

then by the unsatisfiability theorem:

P1 and P2 and P3 and ... and ψ

is unsatisfiable.

e. If Γ ⊨ ¬ψ, then Γ ⊭ ψ.

Using proof by counterexample. If:

Γ = {(ψ and not ψ)} then:

(ψ and not ψ) implies ¬ψ

is valid, because (ψ and not ψ) is unsatisfiable. So:

(ψ and not ψ) implies ψ

is also valid. And so:

Γ ⊨ ψ

is true, which means:

Γ ⊭ ψ

is false.