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Proof step 2 from proof lastN-rest (previous step)

n < # x ⊦ restn(# x.rest - n, x.rest) = restn(# x - n, x), x = []

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%c: ⊦ a + x = a ' + x; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%c-07: ⊦ (a ' + x).rest = x; from specification list%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y