Search for blocks/addresses/...

Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ιι be given.
Assume H0: bij x0 x1 x2.
Apply H0 with bij x1 x0 (inv x0 x2).
Assume H1: and (∀ x3 . prim1 x3 x0prim1 (x2 x3) x1) (∀ x3 . prim1 x3 x0∀ x4 . prim1 x4 x0x2 x3 = x2 x4x3 = x4).
Apply H1 with (∀ x3 . prim1 x3 x1∃ x4 . and (prim1 x4 x0) (x2 x4 = x3))bij x1 x0 (inv x0 x2).
Assume H2: ∀ x3 . prim1 x3 x0prim1 (x2 x3) x1.
Assume H3: ∀ x3 . prim1 x3 x0∀ x4 . prim1 x4 x0x2 x3 = x2 x4x3 = x4.
Assume H4: ∀ x3 . prim1 x3 x1∃ x4 . and (prim1 x4 x0) (x2 x4 = x3).
Claim L5: ...
...
Apply and3I with ∀ x3 . prim1 x3 x1prim1 ((λ x4 . prim0 (λ x5 . and (prim1 x5 x0) (x2 x5 = x4))) x3) x0, ∀ x3 . prim1 x3 x1∀ x4 . prim1 x4 x1(λ x5 . prim0 (λ x6 . and (prim1 x6 x0) (x2 x6 = x5))) x3 = (λ x5 . prim0 (λ x6 . and (prim1 x6 x0) (x2 x6 = x5))) x4x3 = x4, ∀ x3 . prim1 x3 x0∃ x4 . and (prim1 x4 x1) ((λ x5 . prim0 (λ x6 . and (prim1 x6 x0) (x2 x6 = x5))) x4 = x3) leaving 3 subgoals.
Let x3 of type ι be given.
Assume H6: prim1 x3 x1.
Apply L5 with x3, prim1 ((λ x4 . prim0 (λ x5 . and (prim1 x5 x0) (x2 x5 = x4))) x3) x0 leaving 2 subgoals.
The subproof is completed by applying H6.
Assume H7: prim1 (prim0 (λ x4 . and (prim1 x4 x0) (x2 x4 = x3))) x0.
Assume H8: x2 (prim0 (λ x4 . and (prim1 x4 x0) (x2 x4 = x3))) = x3.
The subproof is completed by applying H7.
Let x3 of type ι be given.
Assume H6: prim1 x3 x1.
Let x4 of type ι be given.
Assume H7: prim1 x4 x1.
Assume H8: (λ x5 . prim0 (λ x6 . and (prim1 x6 x0) (x2 x6 = x5))) x3 = (λ x5 . prim0 (λ x6 . and (prim1 x6 x0) (x2 x6 = x5))) x4.
Apply L5 with x3, x3 = x4 leaving 2 subgoals.
The subproof is completed by applying H6.
Assume H9: prim1 ((λ x5 . prim0 (λ x6 . and ... ...)) ...) ....
...
...