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Proof step 6 from proof app-06 (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: x0, a, b abort: F discard: F computed: T

Additional information

Used terms: x0, a, b