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Proof step 7 from proof two (previous step)

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 ⊦ b ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ' + b ' = x + a0 ' + c ' + a1 ' + y0, a ∈ x, ∀ a2, b0, x0, a3, c0. a2 ' + b0 ' = x0 + a3 ' + c0 ' + y0 ∧ (a2 ∈ x0 ∨ a2 = a3 ∨ a2 = c0 ∨ a2 ∈ y0) → x0 = [] ∧ y0 = [] ∧ a2 = a3 ∧ b0 = c0 ⊦ b ∈ x

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ a ∈ b ' ↔ a = b;%a: ⊦ (x + y) + z = x + y + z;%in-16: ⊦ x + a ' = y + b ' + z ∧ a ≠ b → a ∈ z;