Geometry

[cherrycherry]

Suppose the following have been established or assumed
axiom 1: p -> ~y
axiom 2: ~q -> r
Theorem 1: p -> ~z
Theorem 2: x -? either q or z
Theorem 3: r - > either x or y
write a two-column proof of the proposition p -> q

[ewmayer]

Without a definition of your symbology, this is gibberish.

[Wacky]

Moreover, I'm confused by your use of terminology.

It appears to involve "predicate calculus" or "symbolic logic", not anything that I know under the heading of "geometry".

Further, "Theorems" were what we proved.

Since "z" is not mentioned in the axioms, there is no way that if could be derived from them.

Please return to "Square Zero".

[mfgoode]

Originally Posted by Wacky

Moreover, I'm confused by your use of terminology.

It appears to involve "predicate calculus" or "symbolic logic", not anything that I know under the heading of "geometry".

Further, "Theorems" were what we proved.

Since "z" is not mentioned in the axioms, there is no way that if could be derived from them.

Please return to "Square Zero".

I Think you are right Wacky. This is symbolic logic with a language of its own.

I'll have to dig up my books on this symbolism to understand those statements.

I lost interest in Logic when Russell postulated a paradox to some unfortunate's life's work. Than Go'del proved to the effect that by logic one cant sort out the postulates of arithmetic Im writing ad lib without reference so do not expect to be taken up for these statements.

Mally

[Jens K Andersen]

Originally Posted by cherrycherry

Suppose the following have been established or assumed
axiom 1: p -> ~y
axiom 2: ~q -> r
Theorem 1: p -> ~z
Theorem 2: x -? either q or z
Theorem 3: r - > either x or y
write a two-column proof of the proposition p -> q

A possible interpretation:
Ignore "Geometry".
Assume a variable is either true or false.
-> means implies. ~ means not.
"either a or b" means a xor b (exactly one of a, b is true)
Assume -? is a typo for ->
Assume all "axioms" and "Theorems" are true.
Prove p -> q
Ignore "two-column proof"
With the given assumptions, p -> q is provable.

Hint 1: a -> b is equivalent to ~b -> ~a
Hint 2: If you cannot see whether a variable is true or false then examine whether it matters.

[Wacky]

Originally Posted by Jens K Andersen

A possible interpretation:
Ignore "Geometry".
Assume a variable is either true or false.
-> means implies. ~ means not.
"either a or b" means a xor b (exactly one of a, b is true)
Assume -? is a typo for ->
Assume all "axioms" and "Theorems" are true.
Prove p -> q
Ignore "two-column proof"
With the given assumptions, p -> q is provable.

Hint 1: a -> b is equivalent to ~b -> ~a
Hint 2: If you cannot see whether a variable is true or false then examine whether it matters.

Note that it is also true if "either a or b" means "a or b" (at least one is true).

But the point is that someone wanting "help" should not leave the reader guessing. I would also suspect that a portion of the assignment was to make some of the statements of the type that you have provided.

Hint 3: (a implies b) also is equivalent to ( b or (not a))