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Proof step 6 from proof len-01 (previous step)

# x0 = # z0, x0 + y = z0 + y0, ∀ z, y, y0. # z0 = # z ∧ x0 + y = z + y0 → x0 = z ∧ y = y0, ∀ a, x0, y, y0. (a ' + x0) + y = z0 + y0 ∧ # x0 + 1 = # z0 ∧ (∀ z, y, y0. # x0 = # z ∧ x0 + y = z + y0 → x0 = z ∧ y = y0) ∧ (a ∈ z0 ∨ a ∈ y0) → a ' + x0 = z0 ∧ y = y0 ⊦ x0 = z0 ∧ y = y0

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: z0, y, y0 abort: F discard: F computed: T

Additional information

Used terms: z0, y, y0