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Proof step 15 from proof eq (previous step)

x1 + a ' + b ' + y + c ' + z = (x0 + c ' + a ').tail + z0, a0 = (x0 + c ' + a ').head, a0 ∈ x0, ∀ a, b, y, c, z, x0, z0. x1 + a ' + b ' + y + c ' + z = x0 + c ' + a ' + z0 → a = b ∨ a = c ∨ dups(x1 + z) ∨ a ∈ y ∨ a ∈ x1 ∨ a ∈ z ∨ c ∈ x1 ∨ c ∈ z ⊦ x0 + c ' + a ' = [], a = b, a = c, a = a0, c = a0, dups(x1 + z), a ∈ y, a ∈ x1, a ∈ z, c ∈ x1, c ∈ z, a0 ∈ x1, a0 ∈ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
x1 + a ' + b ' + y + c ' + z = (x0 + c ' + a ').tail + z0, (x0 + c ').head ∈ x0, ∀ a, b, y, c, z, x0, z0. x1 + a ' + b ' + y + c ' + z = x0 + c ' + a ' + z0 → a = b ∨ a = c ∨ dups(x1 + z) ∨ a ∈ y ∨ a ∈ x1 ∨ a ∈ z ∨ c ∈ x1 ∨ c ∈ z ⊦ a = c, a = b, a = x0.head, c = x0.head, dups(x1 + z), c ∈ z, c ∈ x1, a ∈ z, a ∈ x1, a ∈ y, x0.head ∈ z, x0.head ∈ x1

Rule argument

Additional information

Used simplifier rules: ⊦ (x + a ' + y).head = (x + a ').head; (s)% ⊦ x + a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x → (x + y).head = x.head; (s)