At the cafe in Kyoto, 毅+明 look at owls。 毅 f(look) var(Owl|various) + 明 f(look)

var(Owl|various)。 Also: 毅 f(pet) var(Cat|Bengal) + 明 f(pet) var(Cat|Bengal) on a

combo ticket, because (f(price) var(Owl) + var(Cat)) < f(price) var(Owl) + f(price)

var(Cat)。 Time=Time+var(x)。 Afterwards the rain, f(rain) var(duration|hour|1)。

Then f(wefie) @ "the giant crab" loc(Kani Doraku|Kyoto) with umbrellas。Then dinner

at Mishima-Tei = f(sukiyaki) var(Beef|Kobe|A5)。 Does 毅+var(joy|y) = 明+var(joy|y)?

Calculate to the nearest dollar。

Time=Time+var(Month|1)。

毅 swims 23% faster than 明。 In other words: 毅 f(swim) > 明 f(swim)

var(speed)*1.23。 毅 waits Time+wait(var(Minute|1)) for 明 to catch up, thinks 明 is

making fun, that 明 f(fun)=f(swim)-var(speed|delay|arbitrary)。 But 明 is only trying to

f(breathe) and f(swim) and 明 var(lungcapacity) < 毅 var(lungcapacity) because 明

var(age|years) = 毅 var(age|years)+24 and past the threshold where it matters。

Determine delay until next 毅 f(swim) || 明 f(swim)。

Time=Time+var(Week|1)。

毅 & 明 share a few strands of acid + a stack of need。 To elaborate: 毅 =

f(undefinedfunction) var(DNA|undefined subvariable)明; 毅 f(need) 明 > 明 f(need) 毅,

but paradox of the equation = 明-毅 > 毅-明。 To solve, let 毅 be 毅, not var(明)。 Let 明

approach 0 as Time=Time+var(x)。 Find (x) first. Bury (x)。