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Proof step 5 from proof lastn-mklist (previous step)

m ≤ n0 + 1, ∀ m, a. m ≤ n0 → lastn(m, mklist(a, n0)) = mklist(a, m) ⊦ lastn(m, mklist(a, n0 + 1)) = mklist(a, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n < # x → lastn(n, x) = lastn(n, x.tail)
Proof step 7:
m < # mklist(a, n0 + 1), m ≤ n0 + 1, ∀ m, a. m ≤ n0 → lastn(m, mklist(a, n0)) = mklist(a, m) ⊦ lastn(m, mklist(a, n0 + 1).tail) = mklist(a, m)

Rule argument

spec: <toplevel> name: lastN-rec-02 seq: ⊦ n < # x → lastn(n, x) = lastn(n, x.tail) subst: [x, n] → [mklist(a, n0 + 1), m] flags: <all paths>

Additional information

Used simplifier rules: lastn-mklist-03: ⊦ lastn(n, mklist(a, n)) = mklist(a, n);% ⊦ # mklist(a, n) = n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ mklist(a, m) = mklist(a, n) ↔ m = n; (s) Used terms: mklist(a, n0 + 1), m