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Proof step 5 from proof len-01 (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 6:
(a ' + x0) + y = (a0 + z0) + y0, 1 + # x0 = #(a0 + z0), ∀ z, y, y0. # x0 = # z ∧ x0 + y = z + y0 → x0 = z ∧ y = y0, ∀ x0, a, y, y0. 1 + # x0 = # z0 ∧ (a ' + x0) + y = z0 + y0 ∧ (∀ z, y, y0. # x0 = # z ∧ x0 + y = z + y0 → x0 = z ∧ y = y0) ∧ (a ∈ z0 ∨ a ∈ y0) → a ' + x0 = z0 ∧ y = y0, a ∈ a0 + z0 ∨ a ∈ y0 ⊦ a ' + x0 = a0 + z0 ∧ y = y0

Rule argument

indvar: z

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic