 By Stefan Bilaniuk

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Suppose M is a structure for L, t is a term of L, and r and s are assignments for M such that r(x) = s(x) for every variable x which occurs in t. Then r(t) = s(t). 8. Suppose M is a structure for L, ϕ is a formula of L, and r and s are assignments for M such that r(x) = s(x) for every variable x which occurs free in ϕ. Then M |= ϕ[r] if and only if M |= ϕ[s]. 9. Suppose M is a structure for L and σ is a sentence of L. Then M |= σ if and only if there is some assignment s : V → |M| for M such that M |= σ[s].

3) Vector spaces. We will need a few additional concepts and facts about formulas of first-order logic later on. First, what are the subformulas of a formula? 10. Define the set of subformulas of a formula ϕ of a first-order language L. For example, if ϕ is (((¬∀v1 (¬ = v1 c7)) → P32 v5v8) → ∀v8(= v8 f53 c0 v1v5 → P21 v8)) in the language L1 , then the set of subformulas of ϕ, S(ϕ), ought to include • = v1c7 , P32 v5v8, = v8f53 c0 v1v5, P21 v8, • (¬ = v1c7 ), (= v8 f53 c0v1 v5 → P21 v8 ), • ∀v1 (¬ = v1c7 ), ∀v8(= v8f53 c0 v1v5 → P21 v8), • (¬∀v1 (¬ = v1 c7)), • (¬∀v1 (¬ = v1 c7)) → P32 v5v8), and • (((¬∀v1 (¬ = v1c7 )) → P32v5 v8) → ∀v8(= v8 f53 c0v1 v5 → P21 v8)) itself.

We can verify that R |= ∀v1 (= v3 · 0v1 →= v30) [s] as follows: R |= ∀v1 (= v3 · 0v1 →= v3 0) [s] ⇐⇒ for all a ∈ |R|, R |= (= v3 · 0v1 →= v3 0) [s(v1|a)] ⇐⇒ for all a ∈ |R|, if R |== v3 · 0v1 [s(v1|a)], then R |== v3 0 [s(v1|a)] ⇐⇒ for all a ∈ |R|, if s(v1|a)(v3) = s(v1 |a)(·0v1), then s(v1|a)(v3) = s(v1|a)(0) ⇐⇒ for all a ∈ |R|, if s(v3) = s(v1|a)(0) · s(v1|a)(v1), then s(v3) = 0 ⇐⇒ for all a ∈ |R|, if s(v3) = 0 · a, then s(v3 ) = 0 ⇐⇒ for all a ∈ |R|, if 4 = 0 · a, then 4 = 0 ⇐⇒ for all a ∈ |R|, if 4 = 0, then 4 = 0 .