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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 ι be given.
Let x3 of type ιιι be given.
Let x4 of type ιιι be given.
Apply explicit_Field_E with x0, x1, x2, x3, x4, explicit_Field_minus x0 x1 x2 x3 x4 x1 = x1.
Assume H0: explicit_Field x0 x1 x2 x3 x4.
Assume H1: ∀ x5 . x5x0∀ x6 . x6x0x3 x5 x6x0.
Assume H2: ∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x3 x5 (x3 x6 x7) = x3 (x3 x5 x6) x7.
Assume H3: ∀ x5 . x5x0∀ x6 . x6x0x3 x5 x6 = x3 x6 x5.
Assume H4: x1x0.
Assume H5: ∀ x5 . x5x0x3 x1 x5 = x5.
Assume H6: ∀ x5 . x5x0∃ x6 . and (x6x0) (x3 x5 x6 = x1).
Assume H7: ∀ x5 . x5x0∀ x6 . x6x0x4 x5 x6x0.
Assume H8: ∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x4 x5 (x4 x6 x7) = x4 (x4 x5 x6) x7.
Assume H9: ∀ x5 . x5x0∀ x6 . x6x0x4 x5 x6 = x4 x6 x5.
Assume H10: x2x0.
Assume H11: x2 = x1∀ x5 : ο . x5.
Assume H12: ∀ x5 . x5x0x4 x2 x5 = x5.
Assume H13: ∀ x5 . x5x0(x5 = x1∀ x6 : ο . x6)∃ x6 . and (x6x0) (x4 x5 x6 = x2).
Assume H14: ∀ x5 . x5x0∀ x6 . x6x0∀ x7 . x7x0x4 x5 (x3 x6 x7) = x3 (x4 x5 x6) (x4 x5 x7).
Apply explicit_Field_plus_cancelL with x0, x1, x2, x3, x4, x1, explicit_Field_minus x0 x1 x2 x3 x4 x1, x1 leaving 5 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H4.
Apply explicit_Field_minus_clos with x0, x1, x2, x3, x4, x1 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H4.
The subproof is completed by applying H4.
Apply H5 with x1, λ x5 x6 . x3 x1 (explicit_Field_minus x0 x1 x2 x3 x4 x1) = x6 leaving 2 subgoals.
The subproof is completed by applying H4.
Apply explicit_Field_minus_R with x0, x1, x2, x3, x4, x1 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H4.