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1d
comment proposition in mathematical logic
Propositions concerning future events are neither true nor false even though often clear as to what they mean. From what I've read from others, Aristotle pointed this out long ago. This is no time reference in "x+2=5". As it stands, it is not specific, and it's not clear exactly what it means. Does it mean "for all x, x+2=5" or "for many x, x+2=5" or "for at least 6 x, x+2=5" or "for a lot of x, x+2=5" or "for very few though not none x, x+2=5" or "for at least one x, x+2=5"?
2d
comment Why exactly are NAND and NOR the only universal binary logic functions?
@RayToal A corollary of that analysis comes as that no formula which just has implication or "implies" symbols qualifies as a contradiction. In other words, if a formula only has implication symbols, then it comes as a true for at least one valuation of the variables.
2d
comment Why exactly are NAND and NOR the only universal binary logic functions?
@RayToal Suppose that B(0, 0)=0 where B is some logic function. Then it follows that where all the inputs equal 0, the formula will have value of 0. Consequently, every formula which just has function B is not a tautology. Tautologies have to come as representable. So, if B(0, 0)=0, then no tautology can get represented, and thus we can only have a tautology when B(0, 0)=1. A similar argument leads us to conclude that if B(1, 1)=1, then no formula with just "B"'s and variables can represent a contradiction. So, if B can represent contradictions and tautologies, then B(0, 0)=1, and B(1, 1)=0.
Apr
14
revised What is the formal proof for distributive law, from the other side of the equation?
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Apr
14
answered What is the formal proof for distributive law, from the other side of the equation?
Apr
14
answered What is the meaning of an algebra?
Apr
14
comment Prove $\vdash (\lnot B \rightarrow \lnot A) \rightarrow (( \lnot B \rightarrow A) \rightarrow B)$
@Tennisman Your citation demonstrated Git Gud's point. It classifies it as a metatheorem. If you're using a metatheorem, then you're not giving a Hilbert style proof, or equivalently formal proof which happens in the object logic. The deduction metatheorem tells you that a formal proof exists (and its meta-proof tells you how to find a formal proof), but it doesn't actually exhibit a formal proof.
Apr
14
answered Construct XNOR with only OR gates
Apr
13
answered Reductio Ad Absurdum Question
Apr
13
answered Automatic theorem prover for proving simple theorems?
Apr
11
comment Why $¬¬\bot \not\in PROP$?
This all said if the author had defined PROP with iii as "if $\phi$ belongs to PROP, then ¬$\phi$ belongs to PROP" no ambiguities similar to how ambiguities would occur if you dropped the parentheses in the second clause. Also, in that case ¬¬⊥ would belong to PROP, but (¬(¬(⊥)) would not belong to PROP.
Apr
9
comment How's my proof?
You could represent the function f such that f(x)=x using $\land$ and $\lor$ by using [x$\land$(x$\lor$y)]. There's a Boolean function "1" such that 1 (x, y)=1 for all x, y, which can represent every tautology. Can you make any tautologies using just $\land$ and $\lor$ and variables?
Apr
9
comment Prove if Tautology, Contradicton, or Neither. Is my proof ok?
You might add more detail to show how you inferred both statements, but otherwise your demonstration seems fine.
Apr
3
comment answer to the $(x+1)(x+2)+(x+3)(x+4)+\cdots+(x+99)(x+100)$?
I don't know about others, but I would find this clearer if we had parentheses around "4i$^2$-2i" to indicate that the summation applies to the all of that.
Apr
3
revised answer to the $(x+1)(x+2)+(x+3)(x+4)+\cdots+(x+99)(x+100)$?
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Apr
3
revised answer to the $(x+1)(x+2)+(x+3)(x+4)+\cdots+(x+99)(x+100)$?
edited body
Apr
3
answered answer to the $(x+1)(x+2)+(x+3)(x+4)+\cdots+(x+99)(x+100)$?
Apr
2
comment When can you perform the same operation on both sides of an equality?
@MarioCarneiro And no, the problem doesn't here doesn't have anything to do with a lack of symmetry suggested by the notation. Consider the equation [x#(y^z)]=[(x#y)^z] where "#" and "^" each indicate some binary operation. Now associativity is a special case of that equation where #=^. We could have a convention that abc means [a#(b$c)]. How has the notation "abc" suggested such a symmetry in such a case?
Apr
2
comment When can you perform the same operation on both sides of an equality?
@MarioCarneiro I didn't claim that suppressing the operation symbol causes the confusion. But, NO, the confusion does not come from suppressing parentheses which indicate the order of operations. If ab=cd, then eab=eab. Now replace ab with cd on the right hand side of "eab=eab" and we obtain eab=ecd. Likewise we could obtain abe=cde. Now I have NOT presupposed that eab means (ea)b NOR have I presupposed that eab means e(ab), it could mean either one, I have no idea which one it means. How does the confusion arise? It arises from suppressing the parentheses implicit in ab=cd.
Apr
2
comment When can you perform the same operation on both sides of an equality?
@MarioCarneiro If abc means [(ab)c], then if ab=cd, then abe=cde, but eab=ecd is false in general. If abc means a(bc), then if ab=cd, then eab=ecd, but abe=cde is false in general. I don't see how either is anymore confusing than suppressing any symbol for a binary operation.