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Proof step 26 from proof put-firstn-restn (previous step)

# y0 = m + 1 ⊦ y0.butlast = firstn(m, y0.butlast), y0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ # x = n → firstn(n, x) = x

Rule argument

spec: <toplevel> name: first-all-01 seq: ⊦ # x = n → firstn(n, x) = x subst: [x, n] → [y0.butlast, m] flags: <all paths>

Additional information

Used simplifier rules: ⊦ x ≠ [] → # x.butlast = # x - 1; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: y0.butlast, m