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Proof step 11 from proof prefixn-shorter (previous step)

prefixn(n, x) = prefixn(n, x0), ∀ n, x1. prefixn(n + 1, x) = prefixn(n + 1, x1) → prefixn(n, x) = prefixn(n, x1) ⊦ n = 0, prefixn(n - 1, x) = prefixn(n - 1, x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
n = m + 1, prefixn(n, x) = prefixn(n, x0), ∀ n, x1. prefixn(n + 1, x) = prefixn(n + 1, x1) → prefixn(n, x) = prefixn(n, x1) ⊦ n = 0, prefixn(m, x) = prefixn(m, x0)

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m