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Computability and Logic

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14.9 Example. Proper use of the two quantifier rules<br />

14.1. SEQUENT CALCULUS 173<br />

(1) Fc ⇒ Fc (R0)<br />

(2) Fc ⇒ Fc, Gc (R1), (1)<br />

(3) Gc ⇒ Gc (R0)<br />

(4) Gc ⇒ Fc, Gc (R1), (3)<br />

(5) Fc ∨ Gc ⇒ Fc, Gc (R4), (2), (4)<br />

(6) Fc ∨ Gc ⇒∃xF x, Gc (R5), (5)<br />

(7) Fc ∨ Gc ⇒∃xF x, ∃xGx (R5), (6)<br />

(8) Fc ∨ Gc ⇒∃xF x ∨∃xGx (R3), (7)<br />

(9) ∃x(F x ∨ Gx) ⇒∃xF x ∨∃xGx (R6), (8)<br />

14.10 Example. Improper use of the second quantifier rule<br />

(1) Fc ⇒ Fc (R0)<br />

(2) Fc, ∼Fc ⇒ (R2b), (1)<br />

(3) ∃xF x, ∼Fc ⇒ (R6), (2)<br />

(4) ∃xF x, ∃x ∼F x ⇒ (R6), (3)<br />

(5) ∃xF x ⇒∼∃x ∼F x (R2b), (4)<br />

(6) ∃xF x ⇒∀xF x (5)<br />

Since ∃xFx does not imply ∀xFx, there must be something wrong in this last<br />

example, either with our rules, or with the way they have been deployed in the<br />

example. In fact, it is the deployment of (R6) at line (3) that is illegitimate. Specifically,<br />

the side condition ‘c not in Ɣ’ in the official statement of the rule is not met, since<br />

the relevant Ɣ in this case would be {∼Fc}, which contains c. Contrast this with a<br />

legitimate application of (R6) as at the last line in the preceding example. Ignoring<br />

the side condition ‘c not in ’ can equally lead to trouble, as in the next example.<br />

(Trouble can equally arise from ignoring the side condition ‘c not in A(x)’, but we<br />

leave it to the reader to provide an example.)<br />

14.11 Example. Improper use of the second quantifier rule<br />

(1) Fc ⇒ Fc (R0)<br />

(2) ∃xF x ⇒ Fc (R6), (1)<br />

(3) ∃xF x, ∼Fc ⇒ (R2b), (2)<br />

(4) ∃xF x, ∃x ∼F x ⇒ (R6), (3)<br />

(5) ∃xF x ⇒∼∃x ∼F x (R2a), (4)<br />

(6) ∃xF x ⇒∀xF x (5)<br />

Finally, let us illustrate the use of the identity rules.<br />

14.12 Example. Reflexivity of identity<br />

(1) c = c ⇒ c = c (R0)<br />

(2) ⇒ c = c (R7), (1)<br />

(3) ∼c = c ⇒ (R2b), (2)<br />

(4) ∃x ∼x = x ⇒ (R6), (3)<br />

(5) ⇒∼∃x ∼x = x (R2a), (4)<br />

(6) ⇒∀xx= x (5)

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