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Proof step 5 from proof firstoc (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 6:
a ∈ x, a ∈ x1, a ∈ x ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), ∀ a, x. a ∈ x ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)) ⊦ a0 ∈ x, ∃ b, y, y0. b ∈ x ∧ a0 ' + x1 = y + b ' + y0 ∧ disj(x, y)

Rule argument

pos: left 3 terms: a, x abort: F discard: F computed: T

Additional information

Used terms: a, x