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Proof step 6 from proof length-from (previous step)

# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → #(x from n1) = n0 ⊦ #(x from n1 + 1) = n0

User applied rule structural induction

And obtained the following new goals

Proof step 7:
∀ n0, n1. # x0 = n0 + n1 + 1 ∧ (∀ x, n0. # x = n0 + n1 → #(x from n1) = n0) → #(x0 from n1 + 1) = n0, #(a + x0) = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → #(x from n1) = n0 ⊦ #(a + x0 from n1 + 1) = n0

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2