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Proof step 9 from proof inapp (previous step)

x0 + a ' + y = (z + a ').tail + y0, z = a0 + y1, (z + a ').head ∈ z, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ x0 ∨ a ∈ y ∨ y = y0 ∧ x0 = z ⊦ a = a0, z = [], a ∈ y, a ∈ x0, y = y0 ∧ a0 ' + x0 = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
x0 + a ' + y = (y1 + a ') + y0, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ x0 ∨ a ∈ y ∨ y = y0 ∧ x0 = z ⊦ a = a0, a ∈ x0, a ∈ y, y = y0 ∧ x0 = y1

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-07: ⊦ (a ' + x).tail = x;%e-01: ⊦ a ' + x ≠ [];%c-08: ⊦ (a ' + x).head = a;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%in-17: ⊦ a ∈ b ' ↔ a = b;%app-01: ⊦ x + y = x + z ↔ y = z;