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Proof step 2 from proof firstn-butlast-intro (previous step)

n < # x ⊦ firstn(n, x) = firstn(n, x.butlast), x = []

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-last inst: name: last-elim seq: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ' subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: first-append: ⊦ m ≤ # x → firstn(m, x + y) = firstn(m, x);%e-02: ⊦ x + a ' ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%len-02: ⊦ # a ' = 0 +1; 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%x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a '; (elim) Used terms: x, a, y