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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ιι be given.
Let x2 of type ιι be given.
Let x3 of type ι be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Let x6 of type ιιι be given.
Let x7 of type ιιι be given.
Assume H0: explicit_Field x0 x3 x4 x6 x7.
Assume H1: ∃ x8 : ι → ι → ο . explicit_Reals {x9 ∈ x0|x1 x9 = x9} x3 x4 x6 x7 x8.
Assume H2: ∀ x8 . x8x0x2 x8{x9 ∈ x0|x1 x9 = x9}.
Assume H3: x5x0.
Assume H4: ∀ x8 . x8x0x1 x8x0.
Assume H5: ∀ x8 . x8x0x2 x8x0.
Assume H6: ∀ x8 . x8x0x8 = x6 (x1 x8) (x7 x5 (x2 x8)).
Assume H7: ∀ x8 . x8x0∀ x9 . x9x0x1 x8 = x1 x9x2 x8 = x2 x9x8 = x9.
Assume H8: x6 (x7 x5 x5) x4 = x3.
Apply and7I with and (and (explicit_Field x0 x3 x4 x6 x7) (∃ x8 : ι → ι → ο . explicit_Reals {x9 ∈ x0|x1 x9 = x9} x3 x4 x6 x7 x8)) (∀ x8 . x8x0x2 x8{x9 ∈ x0|x1 x9 = x9}), x5x0, ∀ x8 . x8x0x1 x8x0, ∀ x8 . x8x0x2 x8x0, ∀ x8 . x8x0x8 = x6 (x1 x8) (x7 x5 (x2 x8)), ∀ x8 . x8x0∀ x9 . x9x0x1 x8 = x1 x9x2 x8 = x2 x9x8 = x9, x6 (x7 x5 x5) x4 = x3 leaving 7 subgoals.
Apply and3I with explicit_Field x0 x3 x4 x6 x7, ∃ x8 : ι → ι → ο . explicit_Reals {x9 ∈ x0|x1 x9 = x9} x3 x4 x6 x7 x8, ∀ x8 . x8x0x2 x8{x9 ∈ x0|x1 x9 = x9} leaving 3 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H8.