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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ιιι be given.
Let x3 of type ιιι be given.
Let x4 of type ιι be given.
Let x5 of type ιι be given.
Assume H0: pack_b_u x0 x2 x4 = pack_b_u x1 x3 x5.
Claim L1: x1 = ap (pack_b_u x0 x2 x4) 0
Apply pack_b_u_0_eq with pack_b_u x0 x2 x4, x1, x3, x5.
The subproof is completed by applying H0.
Claim L2: x0 = x1
Apply L1 with λ x6 x7 . x0 = x7.
The subproof is completed by applying pack_b_u_0_eq2 with x0, x2, x4.
Apply and3I with x0 = x1, ∀ x6 . x6x0∀ x7 . x7x0x2 x6 x7 = x3 x6 x7, ∀ x6 . x6x0x4 x6 = x5 x6 leaving 3 subgoals.
The subproof is completed by applying L2.
Let x6 of type ι be given.
Assume H3: x6x0.
Let x7 of type ι be given.
Assume H4: x7x0.
Apply pack_b_u_1_eq2 with x0, x2, x4, x6, x7, λ x8 x9 . x9 = x3 x6 x7 leaving 3 subgoals.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
Claim L5: x6x1
Apply L2 with λ x8 x9 . x6x8.
The subproof is completed by applying H3.
Claim L6: x7x1
Apply L2 with λ x8 x9 . x7x8.
The subproof is completed by applying H4.
Apply H0 with λ x8 x9 . decode_b (ap x9 1) x6 x7 = x3 x6 x7.
Let x8 of type ιιο be given.
Apply pack_b_u_1_eq2 with x1, x3, x5, x6, x7, λ x9 x10 . x8 x10 x9 leaving 2 subgoals.
The subproof is completed by applying L5.
The subproof is completed by applying L6.
Let x6 of type ι be given.
Assume H3: x6x0.
Apply pack_b_u_2_eq2 with x0, x2, x4, x6, λ x7 x8 . x8 = x5 x6 leaving 2 subgoals.
The subproof is completed by applying H3.
Claim L4: x6x1
Apply L2 with λ x7 x8 . x6x7.
The subproof is completed by applying H3.
Apply H0 with λ x7 x8 . ap (ap x8 2) x6 = x5 x6.
Let x7 of type ιιο be given.
Apply pack_b_u_2_eq2 with x1, x3, x5, x6, λ x8 x9 . x7 x9 x8.
The subproof is completed by applying L4.