Prove the argument:
1. p -> q
2. r \/ s
3. ~s -> ~t
4. ~q \/ s
5. ~s
6. (~p \/ r) -> u
7. w \/ t
therefore u /\ w
Change the disjunctions in 2, 4, and 7, to their equivalent conditionals
by this rule a \/ b <=> ~a -> b
1. p -> q
2. ~r -> s
3. ~s -> ~t
4. q -> s
5. ~s
6. (~p \/ r) -> u
7. ~w -> t
Now write all the equivalent contrapositives of the conditionals:
8. ~q -> p
9. ~s -> r
10. t -> s
11. ~s -> ~q
12. ~s
13. ~u -> ~(~p \/ r)
14. ~t -> w
Start with 5
~s
By 9, we have
~s -> r
By the conditional r -> (r \/ ~p), we have
~s -> r -> (r \/ ~p)
By the commutative law, (r \/ ~p) <=> (~p \/ r), we have:
~s -> r -> (r \/ ~p) <=> (~p \/ r)
By 6, we have
~s -> r -> (r \/ ~p) <=> (~p \/ r) -> u
That is one part of the conjunction conclusion
Start again with 5
~s
By 3, we have
~s -> ~t
By 14, we have
~s -> ~t -> w
So by syllogism we have both parts
of the conjunction conclusion u and w.
Therefore u /\ w
q was not involved so we did not need 1, 4,
or their contrapositives 8, 11.
Edwin
