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pair support double sum2 1. n : 2. m : 3. f : nm 4. x1 : n 5. x2 : n 6. y1 : m 7. y2 : m 8. x1 = x2y1 = y2 9. x:n, y:m. (x = x1 & y = y1) (x = x2 & y = y2) f(x,y) = 0
10. x1 = x2 sum(sum(f(x,y) | y < m) | x < n) = f(x1,y1)+f(x2,y2)
By:
Inst Thm: pairsupport [n;x.sum(f(x,y) | y < m);x1;x2] THEN All ReduceSOAps
9. x:n, y:m. (x = x1 & y = y1) (x = x2 & y = y2) f(x,y) = 0
10. x1 = x2 11. sum(sum(f(x,y) | y < m) | x < n) = sum(f(x1,y) | y < m)+sum(f(x2,y) | y < m)
sum(sum(f(x,y) | y < m) | x < n) = f(x1,y1)+f(x2,y2)
5 steps
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IF YOU CAN SEE THIS go to /sfa/Nuprl/Shared/Xindentation_hack_doc.html