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Proof step 3 from proof two (previous step)

a ' + b ' = x + a0 ' + c ' + (a1 + y0), ∀ a, b, x, a0, c. a ' + b ' = x + a0 ' + c ' + y0 ∧ (a ∈ x ∨ a ∈ a0 ' ∨ a ∈ c ' ∨ a ∈ y0) ∧ (b ∈ x ∨ b ∈ a0 ' ∨ b ∈ c ' ∨ b ∈ y0) → x = [] ∧ y0 = [] ∧ a = a0 ∧ b = c, b ∈ x ∨ b ∈ a0 ' ∨ b ∈ c ' ∨ b ∈ a1 + y0, a ∈ x ∨ a ∈ a0 ' ∨ a ∈ c ' ∨ a ∈ a1 + y0 ⊦ x = [] ∧ a1 + y0 = [] ∧ a = a0 ∧ b = c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ' + b ' = x + a0 ' + c ' + a1 ' + y0, a ∈ x, ∀ a, b, x, a0, c. a ' + b ' = x + a0 ' + c ' + y0 ∧ (a ∈ x ∨ a ∈ a0 ' ∨ a ∈ c ' ∨ a ∈ y0) ∧ (b ∈ x ∨ b ∈ a0 ' ∨ b ∈ c ' ∨ b ∈ y0) → x = [] ∧ y0 = [] ∧ a = a0 ∧ b = c, b ∈ x ∨ b ∈ a0 ' ∨ b ∈ c ' ∨ b ∈ a1 ' ∨ b ∈ y0 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%e-01: ⊦ a ' + x ≠ [];%in-14: ⊦ a ' + x = y → a ∈ y;%in-15: ⊦ x + a ' = y → a ∈ y;%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0);%eq-04: ⊦ x + a ' + y = b ' ↔ x = [] ∧ a = b ∧ y = [];%e: ⊦ a ' ≠ [];