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Proof step 3 from proof inapp-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(a0 ' + x1) + a ' + y = x0 + b ' + z, a0 ∈ x0, ∀ a, y, x0, b, z. x1 + a ' + y = x0 + b ' + z → a ∈ z ∨ b ∈ y ∨ x1 = x0 ∧ a = b ∧ y = z ⊦ b ∈ y, a ∈ z, a0 ' + x1 = x0 ∧ a = b ∧ y = z

Rule argument

Additional information

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