My understanding of these questions is that, given that the three numbered conditions are true, then the proposition (following ∴) is also true. So it appears we are asked to prove or disprove the proposition. I’ve worked out truth tables that will decide whether the propositions in the two questions are correct.
(#1)
(1) R→S
|
R |
S |
R→S |
|
TRUE |
TRUE |
TRUE |
|
TRUE |
FALSE |
FALSE |
|
FALSE |
TRUE |
TRUE |
|
FALSE |
FALSE |
TRUE |
This condition is TRUE for all cases except R TRUE and S FALSE. So we must exclude this combination of R and S in our logic.
(2) T→U is similar to (1), so we exclude T TRUE and U FALSE.
(3) In this condition we limit the truth table to include only the TRUE conditions for (1) and (2).
The table can be represented by 9 12-bit binary numbers where each bit is a table column and each number is a row in the table. 0 represents FALSE, 1 represents TRUE.
The 12 columns from left to right are:
R,~R,S,~S,T,~T,U,~U,~S∨~U,~R∨~S,(~S∨~U)∧(~R∨~S),~R∨~T.
The truth table is:
101010100000
101001100001
101001011001
011010100101
011001100101
011001011111
010110101111
010101101111
010101011111
The last two bits differ in rows 2-5 so the proposition is disproved.
#2)
(1) P→R
Combination P=TRUE, R=FALSE is excluded.
(2) ~P→Q
Combination ~P=TRUE, Q=FALSE is excluded, so P=FALSE, Q=FALSE is excluded.
(3) Q→S
Combination Q=TRUE, S=FALSE is excluded.
The table can be represented by 5 7-bit binary numbers where each bit is a table column and each number is a row in the table. 0 represents FALSE, 1 represents TRUE.
The 12 columns from left to right are:
P,~P,Q,R,~R,S,~R→S
The truth table is:
1011011
1001011
1001001
0111011
0110111
The last bit in each row is 1=TRUE, meaning that for all possible legal combinations the proposition is proven.
Unfortunately there appears to be an abnormality in the system which is preventing me from inserting the truth tables. Also, I get the same error if I try to use comments to show the table. I will try again in a few days. My conclusion from the truth table for #1 is that the proposition is false and #2 is true. The binary representations show this in the absence of the actual truth tables.