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Proof step 11 from proof firstn-firstn (previous step)

m0 < n, n ≤ # y + 1, m0 ≤ # y, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ n = 0, firstn(m0, firstn(n - 1, y)) = firstn(m0, y)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
n = m + 1, m0 < n, n ≤ # y + 1, m0 ≤ # y, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ n = 0, firstn(m0, firstn(m, y)) = firstn(m0, y)

Rule argument

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

Additional information

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