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Proof step 23 from proof put-firstn-restn (previous step)

#(y0.butlast + y0.last ') = m + 1, y0 = y0.butlast + y0.last ', # y0.butlast ≤ m ⊦ y0.butlast + y0.last '[m - # y0.butlast, a] = firstn(m, y0.butlast + y0.last ') + a ', y0.butlast + y0.last ' = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ m ≤ # x → firstn(m, x + y) = firstn(m, x)
Proof step 25:
#(y0.butlast + y0.last ') = m + 1, y0 = y0.butlast + y0.last ', # y0.butlast ≤ m, m ≤ # y0.butlast ⊦ y0.butlast + y0.last ' = [], y0.butlast + y0.last '[m - # y0.butlast, a] = firstn(m, y0.butlast) + a '

Rule argument

spec: <toplevel> name: first-append seq: ⊦ m ≤ # x → firstn(m, x + y) = firstn(m, x) subst: [y, x, m] → [y0.last ', y0.butlast, m] flags: <all paths>

Additional information

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