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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H0: x1 ∈ int.
Let x2 of type ι be given.
Assume H1: x2 ∈ int.
Let x3 of type ι be given.
Assume H2: x3 ∈ int.
Assume H3: divides_int x0 (add_SNo x1 (minus_SNo x2)).
Assume H4: divides_int x0 (add_SNo x2 x3).
Apply H3 with divides_int x0 (add_SNo x1 x3).
Assume H5: and (x0 ∈ int) (add_SNo x1 (minus_SNo x2) ∈ int).
Apply H5 with (∃ x4 . and (x4 ∈ int) (mul_SNo x0 x4 = add_SNo x1 (minus_SNo x2))) ⟶ divides_int x0 (add_SNo x1 x3).
Assume H6: x0 ∈ int.
Assume H7: add_SNo x1 (minus_SNo x2) ∈ int.
Assume H8: ∃ x4 . and (x4 ∈ int) (mul_SNo x0 x4 = add_SNo x1 (minus_SNo x2)).
Apply H8 with divides_int x0 (add_SNo x1 x3).
Let x4 of type ι be given.
Assume H9: (λ x5 . and (x5 ∈ int) (mul_SNo x0 x5 = add_SNo x1 (minus_SNo x2))) x4.
Apply H9 with divides_int x0 (add_SNo x1 x3).
Assume H10: x4 ∈ int.
Assume H11: mul_SNo x0 x4 = add_SNo x1 (minus_SNo x2).
Apply H4 with divides_int x0 (add_SNo x1 x3).
Assume H12: and (x0 ∈ int) (add_SNo x2 x3 ∈ int).
Assume H13: ∃ x5 . and (x5 ∈ int) (mul_SNo x0 x5 = add_SNo x2 x3).
Apply H13 with divides_int x0 (add_SNo x1 x3).
Let x5 of type ι be given.
Assume H14: (λ x6 . and (x6 ∈ int) (mul_SNo x0 x6 = add_SNo x2 x3)) x5.
Apply H14 with divides_int x0 (add_SNo x1 x3).
Assume H15: x5 ∈ int.
Assume H16: mul_SNo x0 x5 = add_SNo x2 x3.
Apply and3I with x0 ∈ int, add_SNo x1 x3 ∈ int, ∃ x6 . and (x6 ∈ int) (mul_SNo x0 x6 = add_SNo x1 x3) leaving 3 subgoals.
The subproof is completed by applying H6.
Apply int_add_SNo with x1, x3 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H2.
Let x6 of type ο be given.
Assume H17: ∀ x7 . and (x7 ∈ int) (mul_SNo x0 x7 = add_SNo x1 x3) ⟶ x6.
Apply H17 with add_SNo x4 x5.
Apply andI with add_SNo x4 x5 ∈ int, mul_SNo x0 (add_SNo x4 x5) = add_SNo x1 x3 leaving 2 subgoals.
Apply int_add_SNo with x4, x5 leaving 2 subgoals.
The subproof is completed by applying H10.
The subproof is completed by applying H15.
Apply mul_SNo_distrL with x0, x4, x5, λ x7 x8 . x8 = add_SNo x1 x3 leaving 4 subgoals.
Apply int_SNo with x0.
The subproof is completed by applying H6.
Apply int_SNo with x4.
The subproof is completed by applying H10.
Apply int_SNo with x5.
The subproof is completed by applying H15.
Apply H11 with λ x7 x8 . add_SNo x8 (mul_SNo x0 x5) = add_SNo x1 x3.
Apply H16 with λ x7 x8 . add_SNo (add_SNo x1 (minus_SNo x2)) x8 = add_SNo x1 ....
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