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logic formula doesn't work in Wolfram Alpha

Posted 3 years ago

(s => !r) && (r && !v => u) && (!t && !r => w or u) && (!x => !u) && (r => x xor s) && (v => t xor r) && (w => v) && (x => t) && (t => !v && r)

Doesn't work with or without a preceding "truth table". Is there a maximum length? I am trying to formalize the following equation:

(if s then (not r)) and ((if r and not v) then u) and ((if not t and not r) then w or u) and (if not x then not u) and (if r then either t or s) and (if v then either t or r) and (if w then v) and (if x then t) and (if t then (not v) and r)

POSTED BY: G Henry
4 Replies
Posted 3 years ago

Good idea! And thanks again!

POSTED BY: G Henry
Posted 3 years ago

I don't know, probably can't really know. Maybe someone with more internal information could provide a more definite answer, but likely not. It was simply an accident that I happened to stumble onto this solution. I have seen a few times in the past where WA seemed to treat some names differently from others. I would watch for x,y,z if you are not expecting those to be generic numeric values. Watch for e,i,n,m. You might watch for other one or two or three character names that tend to have an agreed on definition or use in mathematics or elsewhere. I'm really glad that it worked out for you.

POSTED BY: Bill Nelson
Posted 3 years ago
POSTED BY: G Henry
Posted 3 years ago
POSTED BY: Bill Nelson
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