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Proof step 4 from proof firstoc (previous step)

a ∈ x, ∀ a, x. a ∈ x1 ∧ a ∈ x → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), a = a0 ∨ a ∈ x1 ⊦ ∃ b, y, y0. b ∈ x ∧ a0 ' + x1 = y + b ' + y0 ∧ disj(x, y)

User applied rule cut formula

And obtained the following new goals

Proof step 5:
a = a0, a ∈ x, ∀ a, x. a ∈ x1 ∧ a ∈ x → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), a = a0 ∨ a ∈ x1 ⊦ ∃ b, y, y0. b ∈ x ∧ a0 ' + x1 = y + b ' + y0 ∧ disj(x, y)
Proof step 7:
a ∈ x, ∀ a, x. a ∈ x1 ∧ a ∈ x → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), a = a0 ∨ a ∈ x1 ⊦ a = a0, ∃ b, y, y0. b ∈ x ∧ a0 ' + x1 = y + b ' + y0 ∧ disj(x, y)

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0