How to make bounded inductive proof of a certain general theorem in Group Theory

Question

I am trying to proof by induction that:

Given a group G then forall(x,y,z) / ((y^n)x(z^n) = x) => (Order(G) = n)

I am using bounded induction with the following code

;; Derive order n from a single axiom for groups order n.
;; ((Y^n)X(Z^n) = X )=> (order(G) = n) for 1 < n < 23
(declare-sort S)
(declare-fun e () S)
(declare-fun mult (S S) S)
(declare-fun power (S Int) S)
(assert (forall ((x S)) (= e (power x 0))))
(assert (forall ((x S)) (= x (power x 1))))
(assert (forall ((n Int) (x S))
      (=> (and (>= n 2) (<= n 22))
          (= (power x n) (mult x (power x (- n 1)))))))
(assert (= (mult e e)  e))      
(check-sat)
(define-fun p ((x S) (y S) (z S) (w S) (n Int)) Bool
          (=>   (= (mult (power y n) (mult x (power z n)))
             x) (= (power w n) e) )   )
(assert (forall ((x S) (y S) (z S) (w S)) (p x y z w 2)))
(assert (forall ((x S) (y S) (z S) (w S) (n Int))
      (=> (and (> n 2) (<= n 22))
          (= (p x y z w n) (p x y z w (- n 1))))))

;; Bounded inductive proof.
(assert (not (forall ((x S) (y S) (z S) (w S)) (p x y z w 22))))
(check-sat)

And the output is the expected:

sat 
unsat

Run this code online here

I am proving the theorem for 2 < n < 23. When I try with 2 < n < 24 I am obtaining "timeout".

The question is: How to go beyond n=22 in this proof ?

Answer 1

Using (set-option :qi-eager-threshold 70000) suggested by Leonardo de Moura, Z3 is able to prove the theorem until n = 60000 (locally, online until n = 10000).