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Proof step 11 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(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 ∨ a ∈ y ∨ a ∈ x1 ∨ a ∈ z ∨ c ∈ x1 ∨ c ∈ z ∨ dups(x1 + z) ⊦ c = a0, a = a0, a = c, a = b, dups(x1 + z), a0 ∈ z, a0 ∈ x1, c ∈ z, c ∈ x1, a ∈ z, a ∈ x1, a ∈ y

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0); from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list