Let x0 of type ι be given.
Let x1 of type ι → ι → ο be given.
Assume H0: ∀ x2 x3 . x1 x2 x3 ⟶ x1 x3 x2.
Apply xm with
∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x1 x2 x3),
or (∃ x2 . and (x2 ⊆ x0) (and (atleastp 2 x2) (∀ x3 . x3 ∈ x2 ⟶ ∀ x4 . x4 ∈ x2 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ x1 x3 x4))) (∃ x2 . and (x2 ⊆ x0) (and (atleastp x0 x2) (∀ x3 . x3 ∈ x2 ⟶ ∀ x4 . x4 ∈ x2 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ not (x1 x3 x4)))) leaving 2 subgoals.
Assume H1:
∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x1 x2 x3).
Apply orIR with
∃ x2 . and (x2 ⊆ x0) (and (atleastp 2 x2) (∀ x3 . x3 ∈ x2 ⟶ ∀ x4 . x4 ∈ x2 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ x1 x3 x4)),
∃ x2 . and (x2 ⊆ x0) (and (atleastp x0 x2) (∀ x3 . x3 ∈ x2 ⟶ ∀ x4 . x4 ∈ x2 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ not (x1 x3 x4))).
Let x2 of type ο be given.
Assume H2:
∀ x3 . and (x3 ⊆ x0) (and (atleastp x0 x3) (∀ x4 . x4 ∈ x3 ⟶ ∀ x5 . x5 ∈ x3 ⟶ (x4 = x5 ⟶ ∀ x6 : ο . x6) ⟶ not (x1 x4 x5))) ⟶ x2.
Apply H2 with
x0.
Apply andI with
x0 ⊆ x0,
and (atleastp x0 x0) (∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ not (x1 x3 x4)) leaving 2 subgoals.
The subproof is completed by applying Subq_ref with x0.
Apply andI with
atleastp x0 x0,
∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ not (x1 x3 x4) leaving 2 subgoals.
The subproof is completed by applying atleastp_ref with x0.
The subproof is completed by applying H1.
Assume H1:
not (∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ (x2 = x3 ⟶ ∀ x4 : ο . x4) ⟶ not (x1 x2 x3)).
Apply orIL with
∃ x2 . and (x2 ⊆ x0) (and (atleastp 2 x2) (∀ x3 . x3 ∈ x2 ⟶ ∀ x4 . x4 ∈ x2 ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ x1 x3 x4)),
∃ x2 . and (x2 ⊆ x0) (and (atleastp ... ...) ...).