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Proofgold Proof

pf
Let x0 of type ο be given.
Assume H0: ∀ x1 : ι → ι . (∃ x2 : ι → ι → ι → ι . ∃ x3 x4 : ι → ι . MetaAdjunction_strict (λ x5 . True) HomSet (λ x5 . lam_id x5) (λ x5 x6 x7 x8 x9 . lam_comp x5 x8 x9) struct_e PtdSetHom struct_id struct_comp x1 x2 (λ x5 . ap x5 0) (λ x5 x6 x7 . x7) x3 x4)x0.
Apply H0 with λ x1 . pack_e (setsum 1 x1) 0.
Let x1 of type ο be given.
Assume H1: ∀ x2 : ι → ι → ι → ι . (∃ x3 x4 : ι → ι . MetaAdjunction_strict (λ x5 . True) HomSet (λ x5 . lam_id x5) (λ x5 x6 x7 x8 x9 . lam_comp x5 x8 x9) struct_e PtdSetHom struct_id struct_comp (λ x5 . pack_e (setsum 1 x5) 0) x2 (λ x5 . ap x5 0) (λ x5 x6 x7 . x7) x3 x4)x1.
Apply H1 with λ x2 x3 x4 . lam (setsum 1 x2) (λ x5 . combine_funcs 1 x2 (λ x6 . 0) (λ x6 . Inj1 (ap x4 x6)) x5).
Let x2 of type ο be given.
Assume H2: ∀ x3 : ι → ι . (∃ x4 : ι → ι . MetaAdjunction_strict (λ x5 . True) HomSet (λ x5 . lam_id x5) (λ x5 x6 x7 x8 x9 . lam_comp x5 x8 x9) struct_e PtdSetHom struct_id struct_comp (λ x5 . pack_e (setsum 1 x5) 0) (λ x5 x6 x7 . lam (setsum 1 x5) (λ x8 . combine_funcs 1 x5 (λ x9 . 0) (λ x9 . Inj1 (ap x7 x9)) x8)) (λ x5 . ap x5 0) (λ x5 x6 x7 . x7) x3 x4)x2.
Apply H2 with λ x3 . lam x3 (λ x4 . Inj1 x4).
Let x3 of type ο be given.
Assume H3: ∀ x4 : ι → ι . MetaAdjunction_strict (λ x5 . True) HomSet (λ x5 . lam_id x5) (λ x5 x6 x7 x8 x9 . lam_comp x5 x8 x9) struct_e PtdSetHom struct_id struct_comp (λ x5 . pack_e (setsum 1 x5) 0) (λ x5 x6 x7 . lam (setsum 1 x5) (λ x8 . combine_funcs 1 x5 (λ x9 . 0) (λ x9 . Inj1 (ap x7 x9)) x8)) (λ x5 . ap x5 0) (λ x5 x6 x7 . x7) (λ x5 . lam x5 (λ x6 . Inj1 x6)) x4x3.
Apply H3 with λ x4 . lam (setsum 1 (ap x4 0)) (λ x5 . combine_funcs 1 (ap x4 0) (λ x6 . ap x4 1) (λ x6 . x6) x5).
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Apply unknownprop_712520f713b96d5afd10321cea9a3c978868fc53aa35a29461e902d5b5a4ba79 with λ x4 . True, HomSet, λ x4 . lam_id x4, λ x4 x5 x6 x7 x8 . lam_comp x4 x7 x8, struct_e, PtdSetHom, struct_id, struct_comp, λ x4 . pack_e (setsum 1 x4) 0, λ x4 x5 x6 . lam (setsum 1 x4) (λ x7 . combine_funcs 1 x4 (λ x8 . 0) (λ x8 . Inj1 (ap ... ...)) ...), ..., ..., ..., ... leaving 5 subgoals.
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