SDE, BWAPI owner, hacker.

1. If A is true, then B is true

2. A is true

3. Therefore B is true

This looks like:

1. A ⇒ B

2. A

Conclusion. B

We then prove the following: (A ∧ A ⇒ B) ⇒ B

Assume A ∧ A ⇒ B.

1. A ∧ A ⇒ B

2. A (I2, simplification, from 1)

3. A ⇒ B ^ A (E9, commutative law, from 1)

4. A ⇒ B (I2, simplification, from 3)

5. B (I3, modus ponens, from 2 and 4)

∴(A ∧ A ⇒ B) ⇒ B

This is stuff from my theoretical foundations of computer science book (course code 03-60-231 at the University of Windsor). (EDIT: I accidentally swapped the order of the above but the result will still be the same)

If acts of punishing the innocent were wrong, then God would not have killed all the Egyptian firstborns for the sins of their Pharaoh.

God killed the Egyptian firstborns for the sins of their Pharaoh.

Therefore, punishing the innocent is a righteous act.

Let A = acts of punishing the innocent is a righteous act

Let K = God killed the Egyptian firstborns for the sins of their pharaoh.

We have

1. ~A ⇒ ~K

2. K

Conclusion. A

Prove: (~A ⇒ ~K ∧ K) ⇒ A

Assume ~A ⇒ ~K ∧ K.

1. ~A ⇒ ~K ∧ K

2. ~A ⇒ ~K (I2, simplification, from 1)

3. K ∧ ~A ⇒ ~K (E9, commutative law, from 1)

4. K (I2, simplification, from 3)

5. K ⇒ A (E19, not sure what it's called, from 2)

6. A (I3, modus ponens, from 4 and 5)

∴(~A ⇒ ~K ∧ K) ⇒ A

Again, from my book. I didn't skip any steps and only used the base axioms we were given.

Just thought it would be neat for people to see it mathematically.

EDIT: I think we were allowed to just list them from our individual assumptions, but the first few steps shows that there is no cutting corners.

So the alternative would be,

P1. ~A ⇒ ~K

P2. K

C. A

1. ~A ⇒ ~K (from P1)

2. K (from P2)

3. K ⇒ A (E19, not sure what it's called, from 1)

4. A (I3, modus ponens, from 2 and 3)

∴A

I don't think there can be "loops" using this method.

Post has been edited 3 time(s), last time on May 1 2013, 7:06 pm by Heinermann.