I have been working on school project for a few days and have encountered this problem on predicate logic, which I can't figure out. I am given an argument :
H1: p ∧ q , H2: p → ¬(q ∧ z), H3: r → z , H4: m ∧ p → r
This should imply ¬m.
This is what I have got: 1. H1 and use Simplification to get p.
Take p from step 1 and H2 to get p ∧ (p → ¬(q ∧ z)), then use modus ponens to get ¬(q ∧ z).
Use De Morgan's law on ¬(q ∧ z) to get ¬q ∨ ¬z.
Take q from H1 to get q ∧ (¬q ∨ ¬z), which can be rewritten as q ∧ (q → ¬z), which can be simplified using modus ponens to ¬z.
Take ¬z from 4 and H3 to get ¬z ∧ (r → z). This can be simplified using modus tollens to ¬r.
Rewrite H4 using contraposition, arriving at (¬r → ¬(m ∧ p)).
Using De Morgan's we can adjust the left side of the argument, arriving at (¬r → (¬m ∨ ¬p)).
Combining step 5 and 7 we arrive at ¬r ∧ (¬r → (¬m ∨ ¬p)), which can be simplified using modus ponens to ¬m ∨ ¬p.
And now the step I am a little confused about. We assume that H1 is true, hence p must be true, which implies, that its complement, ¬p, must be false. Thus we can restate the result of 8 and get (¬m ∨ false), which logically gives us ¬m.
So, is step 9 correct reasoning? Or am I missing something? I have got really excited about this, and am very keen to get this right. So if there is something wrong, please let me know, I will try to work out some other way.
Thank you!
Yes, $A$ and $\neg A\vee B$ entails $B$, by rule of disjunctive syllogism
So too $p$ and $\neg p\vee\neg m$ entails $\neg m$.
If you prefer, you may use implication equivalence: $\neg p\vee\neg m$ equates to $p\to\neg m$.
Thence $p$ and $p\to\neg m$ entails $\neg m$ by rule of modus ponens.
(… as you already did in step 4, where you reasoned that $q$ and $\neg q\vee\neg z$ entails $\neg z$.)