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

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, y, x, b, z abort: F discard: F computed: T

Additional information

Used terms: a, y, x, b, z