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Proof step 3 from proof neq-01 (previous step)

∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0, (a0 + y0) + c ' + z = y1 + a ' + x ⊦ a = c, a ∈ z, a ∈ a0 + y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(a0 ' + y0) + c ' + z = y1 + a ' + x, a0 ∈ y1, ∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0 ⊦ a = a0, a = c, a ∈ y0, a ∈ z

Rule argument

Additional information

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