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Proof step 11 from proof lastn-lastn-01 (previous step)

m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0, lastn(n, x).butlast) + lastn(n, x).last ' = lastn(m0, x.butlast) + x.last '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ n ≠ 0 → lastn(n, x) = lastn(n -1, x.butlast) + x.last
Proof step 13:
m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0, (lastn(n -1, x.butlast) + x.last).butlast) + (lastn(n -1, x.butlast) + x.last).last ' = lastn(m0, x.butlast) + x.last ', n = 0

Rule argument

spec: <toplevel> name: lastn-rec seq: ⊦ n ≠ 0 → lastn(n, x) = lastn(n -1, x.butlast) + x.last subst: [x, n] → [x, n] flags: <all paths>

Additional information

Used simplifier rules: lastN-zero: ⊦ lastn(0, x) = [];% ⊦ ¬ n < 0; (s) Used terms: x, n