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Proof step 3 from proof inapp-03 (previous step)

∀ a, z, y, b, y0. x0 + a ' + z = y + b ' + y0 → b ∈ x0 ∨ a ∈ y ∨ a = b ∧ x0 = y ∧ z = y0, (a0 + x0) + a ' + z = y + b ' + y0 ⊦ b ∈ a0 + x0, a ∈ y, a = b ∧ a0 + x0 = y ∧ z = y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(a0 ' + x0) + a ' + z = y + b ' + y0, a0 ∈ y, ∀ a, z, y, b, y0. x0 + a ' + z = y + b ' + y0 → b ∈ x0 ∨ a ∈ y ∨ a = b ∧ x0 = y ∧ z = y0 ⊦ b = a0, b ∈ x0, a ∈ y, a = b ∧ a0 ' + x0 = y ∧ z = y0

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);