Logic 101 (#44): Proof Practice #2, Proof By Contradiction Edition

In this practice problem, we encounter what appears to be a modus ponens target. But when that fails, we switch to proof by contradiction, modus tollens, and a steady stream of DeMorgan Law in the hope it will take us to the promise land.

[PunmasterSTP]

Proof by contradiction? More like "with this educational quality, I am smitten"...by your amazing videos! Thanks again for all of your hard work and generosity.

[MichaEichmann]

My first intuition was that because of premise 3, the antecedent of premise 1 or of premise 2 must be true and thus T is true in either case.
I did:

4. (R v P) v (S v Q) | 3, Disjunction introduction
5. (R v S) v (P v Q) | 4, Associativity & commutativity
6. T v T | 5, Constructive dilemma
7. T | 6, Idempotence

[reniersteytler]

What I got:
4. ~T : Assumption for proof by contradiction
5. ~(P v Q): Modes Tollens (4, 2)
6. ~P ^ ~Q : DeMorgans (5)
7. ~P : Simplification (6)
8. R : Disjunctive Syllogism (7, 3)
9. R v S : Disjunctive Introduction (8)
10. T : Modes Podens (9, 1)
11. ~T ^ T : Conjunction Introduction (11, 4) = Contraction
12 T : Proof by Contradiction (4 - 11)

[Pico]

Isn't it better to use direct proof when possible?
And, in this case, it's really easy, I guess.
4. ((R v P) v S) v Q (3 Disjunction introduction, twice)
5. (R v S) v (P v Q) (4 Associativity)
6. T v T (1, 2, 5 Constructive dilemma)
7. T (6 Idempotence)

Isn't more easy?

[lea]

Another case of proof by contradiction let's see if I got it. In my view it would be something like:

I'll start assuming T is false. Then apply modus tollens to first and second line premises to get negation of both antecedents as a result. Rule of replacement to both of them with De Morgan. I convert them into conjunctions of negatives, i.e., (-R and S) for the first line and (-P and -Q) for the second line. Then I proceed to simplify in order to get -R for the first line, and -P for the second one. Then I introduce a conjunction to get (-R and -P). I will go then to the third line premise and re-write through De Morgan as -(-R and -P). Then I will introduce a new conjunction to arrive to [(-R and -P) and -(-R and -P)]. So there I got the contradictory affirmation. Then I can prove that my assumption was false and as a consequence, if -T is false, T has to be true.

[accountbertolini]

4. -(-R^-P) DeMorgans
5. -R Assumption for Conditional Proof
6. -R => P Material Implication Line 3
7. P Modus Ponens
8. (Q v P) => T Commutative Line 2
9. (-Q => P) => T Material Implication
10. -Q => (P => T) Associative
11. P => T Simplication
12. -R => T Conditional Proof (Lines 5 to 11)

[danguy]

I got there using Proof By Contradiction slightly differently. Is this correct?
4. ~T (Assumption for PBC)
5. ~(R v S) (1, 4 Modus Tollens)
6. ~(P v Q) (2, 4 Modus Tollens)
7. ~R ^ ~S (5 DeMorgan's)
8. ~P ^ ~Q (6 DeMorgan's)
9. ~R (7 Simplification)
10. ~P (8 Simplification)
11. P (3, 9 Disjunctive Syllogism)
12. P ^ ~P (10, 11 Conjunction Introduction)
13. T (4-12 Proof By Contradiction)

[torosalvajebcn]

Is it not posible to use the constructive dilemma?  In that case you just need one step.

[torosalvajebcn]

On line 4 you forgot to write "assumption for".

[TheLukeskywalker]

I think a proof by contraposition would be easier.