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Proof step 2 from proof len-01 (previous step)

# x = # z, x + y = z + y0 ⊦ x = z ∧ y = y0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ z, y, y0. # x0 = # z ∧ x0 + y = z + y0 → x0 = z ∧ y = y0, #(a + x0) = # z, (a + x0) + y = z + y0 ⊦ a + x0 = z ∧ y = y0

Rule argument

indvar: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%l: ⊦ # x = 0 ↔ x = []; from specification list-data