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Proof step 24 from proof nodups-butlastn (previous step)

a ∈ butlastn(# x, a0 ' + x.butlast), ∀ a, n. a ∈ butlastn(n, x) ∧ n ≤ # x → a ∈ x ⊦ a = a0, x = [], a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
n = # x ⊦ butlastn(n, x) = []

Rule argument

spec: inst: name: blastN-all-01 seq: n = # x ⊦ butlastn(n, x) = [] subst: [x, n] → [a0 ' + x.butlast, # x] flags: rewrite, current

Additional information

Used simplifier rules: lb: ⊦ x ≠ [] → # x.butlast = # x - 1;%lb: ⊦ x ≠ [] → # x.butlast = # x - 1;%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%l: ⊦ # x = 0 ↔ x = []; from specification list-data%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%in-empty: ⊦ ¬ a ∈ []; from specification list Used terms: a0 ' + x.butlast, # x