SOLUTION: I need to help on the following proofs Proof 1 B ∧ F �(B ∧ G) ------ �G Proof 2 Goal

SOLUTION: I need to help on the following proofs Proof 1 B ∧ F �(B ∧ G) ------ �G Proof 2 Goal - [A → (B →C)] ↔ [(A → B) → (A &

Question 1011177: I need to help on the following proofs
Proof 1
B ∧ F
�(B ∧ G)
------
�G
Proof 2
Goal - [A → (B →C)] ↔ [(A → B) → (A → C)]
Proof 3
∃x (A(x) ∨ B(x))
∃x A(x) → ∀x (C(x) → B(x))
∃x C(x)
Goal ∃x B(x)
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
I need to help on the following proofs
Proof 1

1. B ∧ F 
2. �(B ∧ G) 
   ------ 
        �G 

3. ~B v ~G          2, DeMorgan
4. B               1, Simplification
5. ~~B             4, Double negation
6. ~G              3,6, Disjunctive syllogism

Proof 2

 
By truth table:

Go through the first time with small letters t and f
Go through the second time with 1's for T and 0's for F 
Go through the third time with + for T and - for F

Goal - [A → (B → C)] ↔ [(A → B) → (A → C)] 
        T 1  T t T   +   T t T  1  T t T
        T 0  T f F   +   T t T  0  T f F
        T 1  F t T   +   T f F  1  T t T
        T 1  F t F   +   T f F  1  T f F
        F 1  T t T   +   F t T  1  F t T
        F 1  T f F   +   F t T  1  F t F
        F 1  F t T   +   F t F  1  F t T
        F 1  F t F   +   F t T  1  F t F

As we see there are only +'s under the ↔ so the 
equivalence holds

Proof 3

I have never studied how to do proofs like the third one.
What is it called?  I'll google it and learn what it's all
about.

∃x (A(x) ∨ B(x))
∃x A(x) → ∀x (C(x) → B(x))
∃x C(x)
Goal ∃x B(x)
Edwin
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SOLUTION: I need to help on the following proofs Proof 1 B ∧ F �(B ∧ G) ------ �G Proof 2 Goal - [A → (B →C)] ↔ [(A → B) → (A &