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Proof step 12 from proof firstn-butlast (previous step)

# x0 = n0 + # z, a0 ' + x0 = firstn(n0, a0 ' + x0) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0), a0 ∈ firstn(n0, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ (a0 ' + firstn(n0, x0)).butlast = firstn(n0, a0 ' + x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 14:
# x0 = n0 + # z, a0 ' + x0 = (a0 ' + firstn(n0 -1, x0)) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0), a0 ∈ a0 ' + firstn(n0 -1, x0) ∨ a0 = a ∨ a0 ∈ z ⊦ (a0 ' + firstn(n0, x0)).butlast = a0 ' + firstn(n0 -1, x0), n0 = 0

Rule argument

spec: <toplevel> name: firstN-rec-01 seq: ⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x) subst: [x, a, n] → [x0, a0, n0] flags: <all paths>

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x + [] = x; (s)% ⊦ a '.butlast = []; (s) Used terms: x0, a0, n0