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Proof step 12 from proof firstoc (previous step)

disj(y, x), disj(x, y), b ∈ x, a ∈ x, a0 ∈ x, ∀ a1, x0. (a1 ∈ y ∨ a1 = b ∨ a1 ∈ y0) ∧ a1 ∈ x0 → (∃ b0, y1, y2. b0 ∈ x0 ∧ y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(x0, y1)), a = b ∨ a ∈ y0 ⊦ a = a0, a ∈ y, b ∈ y, a0 ∈ y, ∃ b0, y1, y2. b0 ∈ x ∧ a0 ' + y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(x, y1)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: a0, [], y + (b + y0) abort: F discard: F computed: F

Additional information

Used simplifier rules: e: ⊦ disj(x, []);% ⊦ a + x = a ' + x; (s)% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = a ' + y ↔ x = y; (s) Used terms: a0, [], y + (b + y0)