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Proof step 7 from proof inapp (previous step)

x0 + a ' + y = (z + a ').tail + y0, a0 = (z + a ').head, a0 ∈ z, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ x0 ∨ a ∈ y ∨ y = y0 ∧ x0 = z ⊦ z + a ' = [], a = a0, a ∈ y, a ∈ x0, y = y0 ∧ a0 ' + x0 = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
x0 + a ' + y = (z + a ').tail + y0, (z + a ').head ∈ z, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ x0 ∨ a ∈ y ∨ y = y0 ∧ x0 = z ⊦ a = z.head, a ∈ x0, a ∈ y, y = y0 ∧ z.head ' + x0 = z

Rule argument

Additional information

Used simplifier rules: first-app: ⊦ a ∈ x → (x + y).head = x.head;%a: ⊦ (x + y) + z = x + y + z;%e-02: ⊦ x + a ' ≠ [];