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Proofgold Proof

pf
Let x0 of type ι be given.
Assume H0: x0setminus omega (Sing 0).
Apply setminusE with omega, Sing 0, x0, ∀ x1 . x1int∃ x2 . and (x2int) (∃ x3 . and (x3x0) (x1 = add_SNo (mul_SNo x2 x0) x3)) leaving 2 subgoals.
The subproof is completed by applying H0.
Assume H1: x0omega.
Assume H2: nIn x0 (Sing 0).
Claim L3: ...
...
Claim L4: ...
...
Apply int_SNo_cases with λ x1 . ∃ x2 . and (x2int) (∃ x3 . and (x3x0) (x1 = add_SNo (mul_SNo x2 x0) x3)) leaving 2 subgoals.
Let x1 of type ι be given.
Assume H5: x1omega.
Apply quotient_remainder_nat with x0, x1, ∃ x2 . and (x2int) (∃ x3 . and (x3x0) (x1 = add_SNo (mul_SNo x2 x0) x3)) leaving 3 subgoals.
The subproof is completed by applying H0.
Apply omega_nat_p with x1.
The subproof is completed by applying H5.
Let x2 of type ι be given.
Assume H6: (λ x3 . and (x3omega) (∃ x4 . and (x4x0) (x1 = add_nat (mul_nat x3 x0) x4))) x2.
Apply H6 with ∃ x3 . and (x3int) (∃ x4 . and (x4x0) (x1 = add_SNo (mul_SNo x3 x0) x4)).
Assume H7: x2omega.
Assume H8: ∃ x3 . and (x3x0) (x1 = add_nat (mul_nat x2 x0) x3).
Apply H8 with ∃ x3 . and (x3int) (∃ x4 . and (x4x0) (x1 = add_SNo (mul_SNo x3 x0) x4)).
Let x3 of type ι be given.
Assume H9: (λ x4 . and (x4x0) (x1 = add_nat (mul_nat x2 x0) x4)) x3.
Apply H9 with ∃ x4 . and (x4int) (∃ x5 . and (x5x0) (x1 = add_SNo (mul_SNo x4 x0) x5)).
Assume H10: x3x0.
Assume H11: x1 = add_nat (mul_nat x2 x0) x3.
Let x4 of type ο be given.
Assume H12: ∀ x5 . and (x5int) (∃ x6 . and (x6x0) (x1 = add_SNo (mul_SNo x5 x0) x6))x4.
Apply H12 with x2.
Apply andI with x2int, ∃ x5 . and (x5x0) (x1 = add_SNo (mul_SNo x2 x0) x5) leaving 2 subgoals.
Apply Subq_omega_int with x2.
The subproof is completed by applying H7.
Let x5 of type ο be given.
Assume H13: ∀ x6 . and (x6x0) (x1 = add_SNo (mul_SNo x2 x0) x6)x5.
Apply H13 with x3.
Apply andI with x3x0, x1 = add_SNo (mul_SNo x2 x0) x3 leaving 2 subgoals.
The subproof is completed by applying H10.
set y6 to be ...
Claim L14: ∀ x7 : ι → ο . x7 y6x7 x1
...
Let x7 of type ιιο be given.
Apply L14 with λ x8 . x7 x8 y6x7 y6 x8.
Assume H15: x7 y6 y6.
The subproof is completed by applying H15.
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