Justification for each simplification step
F = ~(~x & y)
= ~~x | ~y __________________________________________________________________
= x | ~y __________________________________________________________________
= ~y | x __________________________________________________________________
= y => x __________________________________________________________________
Given ~(t0 | t4) => ~e1, (t0 | t4) & ~t1 => e1, e1 and ~t0, prove t4 1: e1 _______________________________________________________________________ 2: ~t0 _______________________________________________________________________ 3: ~(t0 | t4) => ~e1 _______________________________________________________________________ 4: e1 => (t0 | t4) _______________________________________________________________________ 5: t0 | t4 _______________________________________________________________________ 6: ~t0 => t4 _______________________________________________________________________ 7: t4 _______________________________________________________________________
Given a => (b | ~c), a & b, and c, prove b 1: a => (b | ~c) [Given] 2: a & b [Given] 3: c [Given] 4: _____________________________ [And-elimination on 2] 5: _____________________________ [Modus Ponens on 1 and 4] 6: _____________________________ [Commutivity of "or" on 5 ] 7: _____________________________ [Apply (x => y) <=> (~x | y) on 6] 8: _____________________________ [Modus Ponens on 7 and 3]
Done? Great! Suppose we have the following definition regarding tennis:
Definition: a serve is called a let if and only if the ball hits the net and its first bounce is in the service box.Here are some additional facts: There is a sensor that beeps if and only if the ball hits the net. If the ball's first bounce is not in the service box, the umpire yells "out".
Given that during the serve the umpire doesn't yell "out" and the sensor beeps, prove that the serve is a let.