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Proof step 2 from proof sublist-firsteq (previous step)

⊦ firstn(n, x) = a ' + y ↔ n = (n - 1) + 1 ∧ x.head = a ∧ firstn(n - 1, x.tail) = y

User applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)% ⊦ a ' + x ≠ []; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m