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Proof step 3 from proof sublist-lenrec (previous step)

0 + n < # x ⊦ sublist(0, n + 1, x) = sublist(0, n, x) + x[0 + n]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: sublist-lenrec-01 seq: ⊦ n < # x → sublist(0, n + 1, x) = sublist(0, n, x) + x[n] subst: [x, n] → [x, n] flags: <all paths>

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%restN-zero: ⊦ restn(0, x) = x;%firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%restN-zero: ⊦ restn(0, x) = x;% ⊦ x + a = x + a '; (s)% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s) Used terms: x, n