Search for blocks/addresses/...

Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Assume H0: divides_nat x0 x1.
Assume H1: divides_nat x1 x2.
Apply and3E with x0omega, x1omega, ∃ x3 . and (x3omega) (mul_nat x0 x3 = x1), divides_nat x0 x2 leaving 2 subgoals.
The subproof is completed by applying H0.
Assume H2: x0omega.
Assume H3: x1omega.
Assume H4: ∃ x3 . and (x3omega) (mul_nat x0 x3 = x1).
Apply H4 with divides_nat x0 x2.
Let x3 of type ι be given.
Assume H5: (λ x4 . and (x4omega) (mul_nat x0 x4 = x1)) x3.
Apply H5 with divides_nat x0 x2.
Assume H6: x3omega.
Assume H7: mul_nat x0 x3 = x1.
Apply and3E with x1omega, x2omega, ∃ x4 . and (x4omega) (mul_nat x1 x4 = x2), divides_nat x0 x2 leaving 2 subgoals.
The subproof is completed by applying H1.
Assume H8: x1omega.
Assume H9: x2omega.
Assume H10: ∃ x4 . and (x4omega) (mul_nat x1 x4 = x2).
Apply H10 with divides_nat x0 x2.
Let x4 of type ι be given.
Assume H11: (λ x5 . and (x5omega) (mul_nat x1 x5 = x2)) x4.
Apply H11 with divides_nat x0 x2.
Assume H12: x4omega.
Assume H13: mul_nat x1 x4 = x2.
Apply and3I with x0omega, x2omega, ∃ x5 . and (x5omega) (mul_nat x0 x5 = x2) leaving 3 subgoals.
The subproof is completed by applying H2.
The subproof is completed by applying H9.
Let x5 of type ο be given.
Assume H14: ∀ x6 . and (x6omega) (mul_nat x0 x6 = x2)x5.
Apply H14 with mul_nat x3 x4.
Apply andI with mul_nat x3 x4omega, mul_nat x0 (mul_nat x3 x4) = x2 leaving 2 subgoals.
Apply nat_p_omega with mul_nat x3 x4.
Apply mul_nat_p with x3, x4 leaving 2 subgoals.
Apply omega_nat_p with x3.
The subproof is completed by applying H6.
Apply omega_nat_p with x4.
The subproof is completed by applying H12.
Apply mul_nat_asso with x0, x3, x4, λ x6 x7 . x6 = x2 leaving 4 subgoals.
Apply omega_nat_p with x0.
The subproof is completed by applying H2.
Apply omega_nat_p with x3.
The subproof is completed by applying H6.
Apply omega_nat_p with x4.
The subproof is completed by applying H12.
Apply H7 with λ x6 x7 . mul_nat x7 x4 = x2.
The subproof is completed by applying H13.