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Proof step 12 from proof eq (previous step)

(a0 + x1) + a ' + b ' + y + c ' + z = x0 + c ' + a ' + z0, ∀ 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, dups((a0 + x1) + z), c ∈ z, c ∈ a0 + x1, a ∈ z, a ∈ a0 + x1, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(a0 ' + x1) + a ' + b ' + y + c ' + z = x0 + c ' + a ' + z0, 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 ⊦ a = a0, a = c, a = b, c = a0, dups(x1 + z), a0 ∈ z, a0 ∈ x1, c ∈ z, c ∈ x1, a ∈ z, a ∈ x1, a ∈ y

Rule argument

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;% ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0); (ws)% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)