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Proof step 13 from proof lastoc (previous step)

a ∈ x1, a ∈ z, ∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ a0 ∈ z, ∃ b, y1, y0. (b = a0 ∨ b ∈ x1) ∧ y + a0 ' + z = y1 + b ' + y0 ∧ disj(a0 ' + x1, y0)

User applied rule all left

And obtained the following new goals

Proof step 14:
a ∈ x1, a ∈ z, a ∈ x1 ∧ a ∈ z → (∃ b, y, y0. b ∈ x1 ∧ z = y + b ' + y0 ∧ disj(x1, y0)), ∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ a0 ∈ z, ∃ b, y1, y0. (b = a0 ∨ b ∈ x1) ∧ y + a0 ' + z = y1 + b ' + y0 ∧ disj(a0 ' + x1, y0)

Rule argument

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

Additional information

Used terms: a, z