MIT24_241F09_lec18 - Logic - Session 18 Applying our formal...

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Unformatted text preview: Logic - Session 18 Applying our formal semantics Let I be the following interpretation: UD={a,b} F:{<a>} G:{<b>} Show that ( ∀ x)Fx is false on I. <b> ∉ I(F) So for arbitrary d, <den I,d[b/x] (x)> ∉ I(F). So by (2.), d[b/x] doesn’t satisfy Fx on I. So not: for every u ∈ UD, d[u/x] satis¡es Fx on I. So by (8.), not: d satis¡es ( ∀ x)Fx. So not every variable assignment satis¡es ( ∀ x)Fx. So by def. of truth, ( ∀ x)Fx is false on I. Show: ( ∃ x)(Fx ⊃ ( ∀ y)Gy) is true on I ( ∃ x)(Fx ⊃ ( ∀ y)Gy) is true on I iff every d for I satis¡es ( ∃ x)(Fx ⊃ ( ∀ y)Gy) on I. By (9.), d satis¡es ( ∃ x)(Fx ⊃ ( ∀ y)Gy) on I iff for some u ∈ UD, d[u/x] satis¡es (Fx ⊃ ( ∀ y)Gy) on I. By (6.), for some u ∈ UD, d[u/x] satis¡es (Fx ⊃ ( ∀ y)Gy) on I iff for some u ∈ UD, either d[u/x] doesn’t satisfy Fx on I or d[u/x] satis¡es ( ∀ y)Gy on I. Prove the RHS. <b> ∉ I(F) So for arbitrary d, <den I,d[b/x] (x)> ∉ I(F)....
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This note was uploaded on 01/25/2012 for the course PHIL 201H1F taught by Professor Derekallen during the Fall '10 term at University of Toronto.

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MIT24_241F09_lec18 - Logic - Session 18 Applying our formal...

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