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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 ιιι be given.
Let x3 of type ιιι be given.
Assume H0: explicit_Ring x0 x1 x2 x3.
Let x4 of type ι be given.
Assume H1: x4x0.
Claim L2: explicit_Ring_minus x0 x1 x2 x3 x4x0
Apply explicit_Ring_minus_clos with x0, x1, x2, x3, x4 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
Claim L3: explicit_Ring_minus x0 x1 x2 x3 (explicit_Ring_minus x0 x1 x2 x3 x4)x0
Apply explicit_Ring_minus_clos with x0, x1, x2, x3, explicit_Ring_minus x0 x1 x2 x3 x4 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying L2.
Apply explicit_Ring_plus_cancelL with x0, x1, x2, x3, explicit_Ring_minus x0 x1 x2 x3 x4, explicit_Ring_minus x0 x1 x2 x3 (explicit_Ring_minus x0 x1 x2 x3 x4), x4 leaving 5 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying L2.
The subproof is completed by applying L3.
The subproof is completed by applying H1.
Apply explicit_Ring_minus_L with x0, x1, x2, x3, x4, λ x5 x6 . x2 (explicit_Ring_minus x0 x1 x2 x3 x4) (explicit_Ring_minus x0 x1 x2 x3 (explicit_Ring_minus x0 x1 x2 x3 x4)) = x6 leaving 3 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
Apply explicit_Ring_minus_R with x0, x1, x2, x3, explicit_Ring_minus x0 x1 x2 x3 x4 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying L2.