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Proof step 9 from proof lastN-ex (previous step)

# x = n0 + n, x = firstn(n0, x) + restn(n0, x) ⊦ n = 0, ∃ y. x = y + lastn(n, x) ∧ # y = n0, x = firstn(n0, x) + lastn(n, x) ∧ # firstn(n0, x) = n0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: lastn-restn-01 seq: n ≤ # x ⊦ lastn(n, x) = restn(# x - n, x) subst: [x, n] → [x, n] flags: <all paths>

Additional information

Used simplifier rules: first-length: ⊦ # firstn(n, x) = n;%first-length: ⊦ # firstn(n, x) = n;% ⊦ m + n - m = n; (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; (ss)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ ¬ m + n < m; (s) Used terms: x, n