This is what I need to prove formally:

1.∃x Cube(x) ∧ Small(d)
Goal :∃x (Cube(x) ∧ Small(d))

I have already tried different ways, but I still can't prove the goal.

1. ∃x Cube(x) ∧ Small(d)
2. ∃x Cube(x)         ∧Eilm1
3.Small(d)            ∧Eilm1    
4.∃x Small(x)         ∃intro


Could you provide some suggestions for how I might prove that?

  • I'm a little bit confused what you're trying to prove is identical to your premise... since Small(d) is bound, ∃x only ranges over ∃x Cube(x) part even if it is ^ to something else. Did you miswrite what you need to prove?
    – virmaior
    Nov 16 '16 at 23:03
  • My premise is ∃x Cube(x) ∧ Small(d). And my goal is ∃x (Cube(x) ∧ Small(d))
    – wenwen
    Nov 16 '16 at 23:42
  • This is an example where adding extra parenthesis is useful. I presume what you mean is that your premise is ∃x[Cube(x)] ∧ Small(d) and your desired conclusion is ∃x [Cube(x) ∧ Small(d)]?
    – Cort Ammon
    Nov 17 '16 at 1:06
  • yes.my premise is ∃x[Cube(x)] ∧ Small(d) and my desired conclusion is ∃x [Cube(x) ∧ Small(d)]
    – wenwen
    Nov 18 '16 at 0:55

I would do the following:

1. ∃x Cube(x) ∧ Small(d)
2. ∃x Cube(x)      ∧Elim1
3. Small(d)        ∧Elim1    
4. | Cube(z)         A
5. | Cube(z) ^ Small(d) ^Intr 3,4
6. | ∃x(Cube(x) ^ Small(d)) ∃Intr 5
7. ∃x(Cube(x) ^ Small(d)) ∃Elim 2,4-6
  • i don't think step 4 is right .
    – wenwen
    Nov 18 '16 at 0:08
  • @wenwen fixed it. It's been a while since I used this syntax (See lagunita.stanford.edu/c4x/Philosophy/LPL-SP/asset/…)
    – virmaior
    Nov 18 '16 at 1:05
  • But when I type step 5 "Cube(z) ^ Small(d) ^Intr 3,4", the fitch tell me it is not a sentence. How can I do that
    – wenwen
    Nov 18 '16 at 1:12
  • To a large extent, that's fitch being fidly, but you may need to repeat 3 after 4 to be able to ^Intr
    – virmaior
    Nov 18 '16 at 1:34
  • 1
    @wenwen I think in Fitch, z is a constant, so that 'Cube(z)' is an open sentence. If you use a and 'Cube(a)', it should work (even without repeating 3).
    – MarkOxford
    Aug 6 '18 at 21:00

I would start with

1. ∃x Cube(x) ∧ Small(d)
2. Small(d)
3. ∃x Cube(x)

Which is true because both sides of a true expression must also be true.

I'd then modify 3 to :

4. ∃x [Cube(x)∧T]

Because you can always intersect something with True without changing its value. I can then substitute one true expression(2) for another true expression (T) because they both have the same truth value

5. ∃x [Cube(x)∧Small(d)]
  • what is rule for the step4 ? I have a little confuse
    – wenwen
    Nov 18 '16 at 1:34
  • I don't know what name it would be given, but it's easy to prove with a basic truth table. For any P and Q, Q→(P↔P∧Q) . As long as Q is true, you can add conjunctions (and) with it as much as you like without changing the truth value of the expression. In my case, Q is the very trivial T, which leads to P↔P∧T. It's like adding 0 or multiplying by 1.
    – Cort Ammon
    Nov 18 '16 at 2:08

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