mamert comments on The Parable of the Dagger

[mamert]

Bx is true if box x has gold, false if frog. one contains frog, other gold → B1 == ~B2. only one inscription is true → Bf == ~Bt

We know:

B2 && Bf || Bt && B1 (I1)

B2 && Bt || B1 && Bt (I2)

Bt == B1 && Bf == B2 && I1 && ~I2 || Bf == B1 && Bt == B2 && ~I1 && I2 # only one inscription is true

From this:

((B2 && B2 || B1 && B1) && ~(B2 && B1 || B1 && B1)) || (~(B2 && B1 || B2 && B1) && (B2 && B2 || B1 && B2))

((B2 || B1) && ~(false || B1)) || (~(false || false) && (B2 || false))

(true && (true && B2)) || ((true && true) && B2)

B2 || B2

B2 # so, Box 2 contains gold