Prove: [ P -> Q AND R -> S AND (P OR R) ] -> (Q OR S) -> NOT, --- 1. P -> Q ___ hypothesis 2. R -> S ___ hypothesis 3. P OR R ___ hypothesis 4. ~P OR Q ___ implication from hyp 1. 5. ~R OR S ___ implication from hyp 2 6. ~P OR Q OR S ___ addition to 4. 7. ~R OR Q OR S ___ addition to 5. 8. Let T == (Q OR S) ___ substitution 9. (~P OR T) AND (~R OR T) ___ Conjunction 6,7 10. T OR (~P AND ~R) ___ Distribution from 9 11. T OR ~(P OR R) ___ De Morgan's theorem 12. Let M == (P OR R) ___ substitution 13. (T OR ~M) AND M ___ conjunction 11, hyp 3 From there, you can use distribution to get (T AND M) OR (~M AND M). The contradiction goes away leaving you with T AND M, which can simplify to T.
5 points for a try in rugby union
Bergerac - 1981 S-P-A-R-T-A- 5-6 is rated/received certificates of: UK:12 (video rating) (2007)
Five points for a try in rugby (Union).
T= Temperature P= Pulse R= Respiration
p a t t e r n
u = p r t r = u / p t
P=s r t , so, s= P/(st)
5 Points for a Try in Rugby Union (UK)
The answer depends on whether you are dealing with simple interest of compound interest. Suppose P = Principle R = Rate (in % per annum) T = Time (in years) I = Interest Then for simple interest: I = P*R*T/100 so that P = 100*I/(R*T) For compound interest P+I = P*(1+R/100)T so that P = I/[(1+R/100)T - 1]
If the question is "5 P of I T?", the answer is "5 parts of intelligence tests".
J. R. P. McKenzie has written: 'T. Mann \\' 'T. Mann \\'