Originally Posted by

**Alea** Given (p ⇒ q) and (r ⇒ s), use the Fitch System to prove (p ∨ r ⇒ q ∨ s).

p => q - premise

r => s - premise

Goal - p | r => q | s

Code:

1. p => q Assumption
2. r => s Assumption
3. p | r Assumption
4. p Assumption
5. q 1, 4: =>E
6. q | s 5: |I
7. r Assumption
8. s 2, 7: =>E
9. q | s 8: |I
10. q | s 3, 4-6, 7-9: |E
11. p | r => q | s 3-10: =>I

Originally Posted by

**Alea** Given (¬*p* ⇒ *q*) and (*q* ⇒ *r*), use the Fitch System to prove ((¬*p* ⇒ ¬*r*) ⇒ *p*).

~p => q - premise

q => r - premise

Goal: (~p => ~r) => p

This one requires double-negation elimination rule. The most one can derive without it is ~~p from the contraposition of ~p => q.

Assume ~p => ~r and ~p. Derive ~r, q and r by =>E. Then r and ~r give a contradiction, so ~~p. From there double-negation elimination gives p.

Rules dealing with negation may vary from one version of Fitch calculus to another.