MIT24_241F09_lec18

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

## 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.

### Page1 / 8

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online