Logic question (conjunction of implications)

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To derive "x&z implies y&w" from "x implies y" and "z implies w" in classical logic, one can utilize logical equivalences and manipulations involving conjunctions and disjunctions. The transformation relies on understanding that "x implies y" is equivalent to "(not x) or y" and similarly for "z implies w." The desired implication can be shown through distributive laws and DeMorgan's Laws, leading to the equivalence of "not (x and z) or (y and w)." For those unfamiliar with these concepts, using truth tables or studying resources on logical operators may be beneficial. Mastery of these logical principles is essential for completing the proof.
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hi all,

I'm no logician but am interested in sorting out this problem.

Say you've shown that

1. x implies y
and
2. z implies w

what steps/assumptions are required, in classical logic, to get from 1&2 to:

3. x&z implies y&w

Do the steps require some sort of separability assumption, or something of the sort?

Thanks!
 
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"x implies y" is logically equivalent to "(not x) or y"
"z impllies w" is logically equivalent to "(not z) or w".

The statement you want to prove is logically equivalent to
"not (x and z) or (y and w)"

Are you skilled enough at manipulating "and","or" and "not" to do the proof?

If not, you could use a truth table.
 
thanks very much stephen!

I am not skilled at all with manipulating operators, so cannot do the proof myself.

Can you suggest a website or textbook I could look at that would help?
 
I don't know of a good website, off hand. Perhaps another forum member will.

An outline of what you need to know is

Distributive laws:

"A and (B or C)" is equivalent to "(A and B) or (A and C)"
"A or (B and C)" is equivalent to "(A or B) and (A or C)".


DeMorgan's Laws:

"not (A and B)" is equivalent to "(not A) or (not B)"
"not (A or B)" is equivalent to "(not A) and (not B)"

Double Negation

"not (not A)" is equivalent to "A"


Material Implication
"if A then B" is equivalent to "(not A) or B"

Or you can search for "Truth Tables" and find a explanation of how to do them.
 
ok thanks very much
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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