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I figured out the symbolization, but I am unable to prove it since my professor does not let us use the EI (existential instantiation) rule. I can see that I have use the =E (substitution) rule if I have...

x = b, Bxc.

But I am not sure on how to get Bxc from the existential quantifier without having to use EI.

Ib ∧ Wb ∧ ∀x((Ix ∧ Wx) → x = b) , ∃x(Ix ∧ Wx ∧ Bxc ∧ ∀y((Wy ∧ Byc) → x = y)) ⊢ Bbc

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  • The second premise says "there is a unique objcet that is Ix and Wx and Bxc": call it d. Thus, from first premise we have : Id and Wd and d=b. Thus, from Bdc it follows : Bbc. Dec 2, 2016 at 19:19
  • @MauroALLEGRANZA Yeah that's what I was going to do, except instead of using d I would've used a different letter. I just am not sure how to get the Bdc from the second premise.
    – K.Wong
    Dec 2, 2016 at 20:04

1 Answer 1

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{1}      1.   Ib & Wb & ∀x[(Ix & Wx) → x=b]                 Prem.
{2}      2.   Ǝx[Ix & Wx & Bxc & ∀y[(Wy & Byc) → x=y]]      Prem.
{1}      3.   ∀x[(Ix & Wx) → x=b]                           1 &E
{4}      4.   Id & Wd & Bdc & ∀y[(Wy & Byc) → d=y]          Assum. TD(d)
{4}      5.   Id & Wd                                       4 &E
{1}      6.   (Id & Wd) → d=b                               3 UE
{1,4}    7.   d=b                                           5,6 MP
{4}      8.   Bdc                                           4 &E
{1,4}    9.   Bbc                                           7,8 =E
{1,2}    10.  Bbc                                           2,4,9 EE
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