1
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(p ⇒ q) ∧ (¬p ⇒ ¬q ∧r)

(q ∨r) ⇒ s

s ⇒ t

Therefore,

t

my efforts.

1 (p ⇒ q) ∧ (¬p ⇒ ¬q ∧r premise

2 (q ∨r) ⇒ s premise

3 s ⇒ t premise

4 p assumption

5 q from 1 & 4 ⇒ elim (modus ponens)

6 q∨r from 2 & 5, v Intro

7 s from 2 & 6, ⇒ elim (modus ponens)

8 t from 3 & 7, ⇒ elim (modus ponens)

9 ¬p assumption

10 ¬q ∧r from 1 & 9 ⇒ elim (modus ponens)

11 r from 10 ∧ Elim

12 q∨r from 11 ∨intro

13 s from 2 & 12, ⇒ elim (modus ponens)

14 t from 3 & 13, ⇒ elim (modus ponens)

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  • $\begingroup$ What is your question? And what do you mean with "each step"? The second line does not follow from the first line, no does the third line follow from the second etc. $\endgroup$ – Hagen von Eitzen Apr 3 '17 at 18:01
  • $\begingroup$ Already asked as validity-of-deductive-proof. $\endgroup$ – Mauro ALLEGRANZA Apr 3 '17 at 18:02
  • $\begingroup$ Hi Tim , apologies for the confusion, the first three lines are my premises, or what i am lead to believe, i need to derive to the conclusion t. I am learning this topic at uni, maths is not really my strongest point. $\endgroup$ – user432124 Apr 3 '17 at 18:04
  • $\begingroup$ @MauroALLEGRANZA thanks, just to confirm is that the correct answer? $\endgroup$ – user432124 Apr 3 '17 at 18:07
  • $\begingroup$ hi tim. i have now provided my working out, i know its wrong. $\endgroup$ – user432124 Apr 3 '17 at 18:14
2
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1$\;(p \to q) \land (\lnot p \to (¬q \land r))$ $\quad$ premise

2$\;(q\lor r)\to s$ $\quad$ premise

3$\;s\to t$. $\quad $ premise


4 $\;(p\to q)\quad$ (1, $\land$-elim)

5 $\;\lnot p \to (\lnot q \land r)\quad (1, \land$-elim)

6 $\;p \lor \lnot p\quad$ (tautology: Assuming Law of the excluded middle)

7 $\qquad$ Assume $p.\;\;$ (assumption)

8 $\qquad\qquad q\quad $ from $4, 7$, elim modus ponens

9$\qquad\qquad q\lor r\quad $ from 8 $\lor$-intro

10 $\qquad\qquad s\quad$ from 2&9 elim modus ponens

11 $\qquad\qquad t\quad$ from 3&10 elim modus ponens

12 $\qquad p\to t\quad$ from $7-11$, $\to$-intro

10 $\qquad$ Assume $\lnot p\quad $ Assumption

11 $\qquad\qquad \lnot q\land r\quad$ 5, 10, modus ponens

12 $\qquad\qquad r\quad$ 11, $\land$-elim

13 $ \qquad \qquad q\lor r\quad $ 12 $\,\lor$-Intro

14$\qquad\qquad s\quad $ from 2, 13, modus ponens

15$\qquad\qquad t\quad$ from 3, 14, modus ponens

16 $\qquad\lnot p \to t\quad $ from 10-15, conditional introduction.

17 $\;(p\lor \lnot p)\to t$

18 $\;t$

Now, you can refer to your text and notes, and the rules of inference you've learned, to justify each step.

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  • $\begingroup$ Let me know if you get stuck on the reason for any of the above steps. $\endgroup$ – Namaste Apr 3 '17 at 18:26
  • $\begingroup$ hi could you possibly label each of the steps, i know its a long stretch, i know some of laws applied on various steps but i'm not 100% sure $\endgroup$ – user432124 Apr 3 '17 at 18:31
  • $\begingroup$ Yes, you've got it! Although 4 in my sketch is the assumption p, so 1+4 $\to 5$ via modus ponens. $\endgroup$ – Namaste Apr 3 '17 at 18:36
  • $\begingroup$ on line 5 you used v intro? $\endgroup$ – user432124 Apr 3 '17 at 18:36
  • $\begingroup$ would it possible to edit your answer and leave the laws applied on each line? $\endgroup$ – user432124 Apr 3 '17 at 18:39

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