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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιιο be given.
Let x2 of type ιιο be given.
Let x3 of type ιιο be given.
Let x4 of type ιιο be given.
Let x5 of type ιο be given.
Let x6 of type ιο be given.
Assume H0: ∀ x7 . x7x0∀ x8 . x8x0iff (x1 x7 x8) (x2 x7 x8).
Assume H1: ∀ x7 . x7x0∀ x8 . x8x0iff (x3 x7 x8) (x4 x7 x8).
Assume H2: ∀ x7 . x7x0iff (x5 x7) (x6 x7).
Claim L3: ...
...
Apply L3 with λ x7 x8 . lam 4 (λ x9 . If_i (x9 = 0) x0 (If_i (x9 = 1) (encode_r x0 x1) (If_i (x9 = 2) (encode_r x0 x3) (Sep x0 x5)))) = lam 4 (λ x9 . If_i (x9 = 0) x0 (If_i (x9 = 1) x7 (If_i (x9 = 2) (encode_r x0 x4) (Sep x0 x6)))).
Claim L4: encode_r x0 x3 = encode_r x0 x4
Apply encode_r_ext with x0, x3, ....
...
Apply L4 with λ x7 x8 . lam 4 (λ x9 . If_i (x9 = 0) x0 (If_i (x9 = 1) (encode_r x0 x1) (If_i (x9 = 2) (encode_r x0 x3) (Sep x0 x5)))) = lam 4 (λ x9 . If_i (x9 = 0) x0 (If_i (x9 = 1) (encode_r x0 x1) (If_i (x9 = 2) x7 (Sep x0 x6)))).
Claim L5: Sep x0 x5 = Sep x0 x6
Apply encode_p_ext with x0, x5, x6.
The subproof is completed by applying H2.
Apply L5 with λ x7 x8 . lam 4 (λ x9 . If_i (x9 = 0) x0 (If_i (x9 = 1) (encode_r x0 x1) (If_i (x9 = 2) (encode_r x0 x3) (Sep x0 x5)))) = lam 4 (λ x9 . If_i (x9 = 0) x0 (If_i (x9 = 1) (encode_r x0 x1) (If_i (x9 = 2) (encode_r x0 x3) x7))).
Let x7 of type ιιο be given.
Assume H6: x7 (lam 4 (λ x8 . If_i (x8 = 0) x0 (If_i (x8 = 1) (encode_r x0 x1) (If_i (x8 = 2) (encode_r x0 x3) (Sep x0 x5))))) (lam 4 (λ x8 . If_i (x8 = 0) x0 (If_i (x8 = 1) (encode_r x0 x1) (If_i (x8 = 2) (encode_r x0 x3) (Sep x0 x5))))).
The subproof is completed by applying H6.