This page of questions is not complete. It is, however, to the best of my knowledge, consistent, although I can not prove it to be so.
-p--p-q----
a theorem of the pq-system? If so, give a derivation
(proof). If not, why not? And why might some humans think that it is?
xp-qx
? That is, what is the change of interpretation?
G(0) = 0 G(n+1) = n+1 - G(G(n))
M(0) = 0 F(0) = 1 F(n+1) = n+1 - M(F(n)) M(n+1) = n+1 - F(M(n))
Q(1) = 1 Q(2) = 1 Q(n+2) = Q(n+2 - Q(n+1)) + Q(n+2 - Q(n))
$\forall b \, \exists m \, \exists n\:T(m,b,n)$ | maximal monkey madness |
$\exists m \, \exists n \, \forall b\:T(m,b,n)$ | monkey rugby |
$\exists m \, \exists b \, \exists n\:T(m,b,n)$ | leave no banana untossed |
$\exists m \, \exists b \, \forall n\:T(m,b,n)$ | one monkey really likes another |
$\exists m \, \forall b \, \exists n\:T(m,b,n)$ | a monkey plays zookeeper at lunchtime |
$\exists m \, \forall b \, \forall n\:T(m,b,n)$ | a monkey shows its banana to everyone |
$\forall m \, \exists b \, \exists n\:T(m,b,n)$ | a banana up in the air |
$\forall m \, \exists b \, \forall n\:T(m,b,n)$ | every monkey tosses every banana |
$\forall m \, \forall b \, \exists n\:T(m,b,n)$ | monkey quarterback practices tossing all the bananas, all monkeys catch |
$\forall m \, \forall b \, \forall n\:T(m,b,n)$ | delivery service monkey delivers all the bananas |
$\exists m \, \forall n \, \exists b\:T(m,b,n)$ | everyone, toss a banana to someone! |
$\exists b \, \forall m \, \forall n\:T(m,b,n)$ | monkey show and tell: every monkey shows their special banana to everyone |
∀a:∃b:b=Sa
to
∃b:b=Sb
.
∃a:a=SSSSa
∀a:(a+S0)=Sa
∀a:(S0+a)=Sa
∀a:a=a
∀a:(a•S0)=a
times
method.