logic homework 5
homework on formal proofs, using logic formulas, deduction theorem
prove/disprove:
premises:
$\lnot (p \lor r)$
theres no way forward.. imma try a counter example
$p \lor r$ has to be true
$\lnot p \lor q$ has to be true
$\lnot r \lor p$ has to be true
pick the combination $p=T,q=T,r=T,s=T$
$\lnot p \lor q \equiv F \lor T \equiv T$
$\lnot q \lor (r \land s) \equiv F \lor (T \land T) \equiv F \lor T \equiv T$
$\lnot r \lor p \equiv F \lor T \equiv T$
$\lnot (p \lor r) \equiv \lnot (T \lor T) \equiv \lnot T \equiv F$
we provided an example where the premises are true but the conclusion isnt, so indeed the argument is invalid
premises:
- $\lnot p \lor q$
- $\lnot q \lor (r \land s)$
- $\lnot r \lor p$
$\lnot (p \lor r)$
theres no way forward.. imma try a counter example
$p \lor r$ has to be true
$\lnot p \lor q$ has to be true
$\lnot r \lor p$ has to be true
pick the combination $p=T,q=T,r=T,s=T$
$\lnot p \lor q \equiv F \lor T \equiv T$
$\lnot q \lor (r \land s) \equiv F \lor (T \land T) \equiv F \lor T \equiv T$
$\lnot r \lor p \equiv F \lor T \equiv T$
$\lnot (p \lor r) \equiv \lnot (T \lor T) \equiv \lnot T \equiv F$
we provided an example where the premises are true but the conclusion isnt, so indeed the argument is invalid
prove/disprove:
premises:
$(p \lor q) \to r$
couldnt make it any further so im gonna propose a counter example
we need an example where the permises are true but the conclusion isnt, the conclusion is false when $r$ is false but $p \lor q$ is true, and according to the first line in the incomplete proof, which is true, if $r$ is false then $p \land q$ must be false aswell, so we pick the combination $p=F,q=T,r=F$ which satisfies these conditions
$p \to (q \to r)$ becomes $F \to (T \to F) \equiv F \to F \equiv T$, check
$q \to (p \to r)$ becomes $T \to (F \to F) \equiv T \to T \equiv T$, check
$(p \lor q) \to r$ becomes $(F \lor T) \to F \equiv T \to F$ which is false, so the conclusion isnt true which means the argument is false
premises:
- $p \to (q \to r)$
- $q \to (p \to r)$
$(p \lor q) \to r$
couldnt make it any further so im gonna propose a counter example
we need an example where the permises are true but the conclusion isnt, the conclusion is false when $r$ is false but $p \lor q$ is true, and according to the first line in the incomplete proof, which is true, if $r$ is false then $p \land q$ must be false aswell, so we pick the combination $p=F,q=T,r=F$ which satisfies these conditions
$p \to (q \to r)$ becomes $F \to (T \to F) \equiv F \to F \equiv T$, check
$q \to (p \to r)$ becomes $T \to (F \to F) \equiv T \to T \equiv T$, check
$(p \lor q) \to r$ becomes $(F \lor T) \to F \equiv T \to F$ which is false, so the conclusion isnt true which means the argument is false
prove/disprove:
premises:
$(a \lor c) \to (b \land d)$
couldnt continue, will try providing a counter example
$a \lor c$ has to be true
$b \land d$ has to be false
we pick $a=T,p=F,b=T,c=F,d=F$
$(a \lor p) \to b \equiv (F \lor F) \to T \equiv T$
$c \to (d \land p) \equiv F \to (F \land F) \equiv F \to F \equiv T$
$(a \lor c) \to (b \land d) \equiv (T \lor F) \to (T \land F) \equiv T \to F \equiv F$
we provided an example where the premises are true but the conclusion isnt, so indeed the argument is invalid
premises:
- $(a \lor p) \to b$
- $c \to (d \land p)$
$(a \lor c) \to (b \land d)$
couldnt continue, will try providing a counter example
$a \lor c$ has to be true
$b \land d$ has to be false
we pick $a=T,p=F,b=T,c=F,d=F$
$(a \lor p) \to b \equiv (F \lor F) \to T \equiv T$
$c \to (d \land p) \equiv F \to (F \land F) \equiv F \to F \equiv T$
$(a \lor c) \to (b \land d) \equiv (T \lor F) \to (T \land F) \equiv T \to F \equiv F$
we provided an example where the premises are true but the conclusion isnt, so indeed the argument is invalid
prove/disprove:
premises:
$\lnot a \to \lnot (c \lor \lnot d)$
premises:
- $\lnot b \to \lnot (\lnot g \to c)$
- $b \to a$
- $\lnot b \to e$
- $\lnot e \lor d$
$\lnot a \to \lnot (c \lor \lnot d)$
prove/disprove:
premises:
$(q \lor r) \to (p \land r)$
premises:
- $p \to q$
- $q \to (r \land s)$
- $r \to p$
$(q \lor r) \to (p \land r)$
prove/disprove:
premises:
$(s \lor r) \to p$
couldnt continue, will try providing a counter example
restrictions for the counter example:
so $p$ is false, $q$ is true, $r$ is true, $s$ is false, give us:
$p \lor q \equiv F \lor T \equiv T$ check
$(p \land s) \to \lnot p \equiv (F \land F) \to T \equiv T$
$(p \land \lnot s) \lor r \equiv (F \land T) \lor T \equiv F \lor T \equiv T$
$(s \lor r) \to p \equiv (F \lor T) \to F \equiv T \to F \equiv F$
we provided an example where the premises are true but the conclusion isnt, so indeed the argument is invalid
premises:
- $p \lor q$
- $(q \land s) \to \lnot p$
- $(p \land \lnot s) \lor r$
$(s \lor r) \to p$
couldnt continue, will try providing a counter example
restrictions for the counter example:
- $s \lor r$ has to be true
- $p$ has to be false
- $p \lor q$ has to be true which means $q$ has to be true
- $r$ has to be true because of $P3$ and $p$ being false
so $p$ is false, $q$ is true, $r$ is true, $s$ is false, give us:
$p \lor q \equiv F \lor T \equiv T$ check
$(p \land s) \to \lnot p \equiv (F \land F) \to T \equiv T$
$(p \land \lnot s) \lor r \equiv (F \land T) \lor T \equiv F \lor T \equiv T$
$(s \lor r) \to p \equiv (F \lor T) \to F \equiv T \to F \equiv F$
we provided an example where the premises are true but the conclusion isnt, so indeed the argument is invalid