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Proof step 4 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 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a ' + b ' = x + a0 ' + c ' + a1 ' + y0, a ∈ x, b ∈ 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 ⊦
Proof step 7:
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

Rule argument

fma: b ∈ x

Additional information

Used formulas: b ∈ x