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Proof step 18 from proof sublist-move (previous step)

m1 ≤ # x0, m ≤ m1 + 1, m ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ m = 0, a ' + firstn(m - 1, firstn(m1, x0)) = firstn(m, a ' + x0)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m2] flags:

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%firstn-firstn: ⊦ m ≤ n ∧ n ≤ # x → firstn(m, firstn(n, x)) = firstn(m, x);%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%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%c: ⊦ a + x = a ' + x; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m2