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Proof step 8 from proof lastn-lastn-01 (previous step)

m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0 + 1, lastn(n, x)) = lastn(m0, x.butlast) + x.last '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ n ≠ 0 → lastn(n, x) = lastn(n -1, x.butlast) + x.last
Proof step 10:
m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn((m0 + 1)-1, lastn(n, x).butlast) + lastn(n, x).last = lastn(m0, x.butlast) + x.last ', m0 + 1 = 0

Rule argument

spec: <toplevel> name: lastn-rec seq: ⊦ n ≠ 0 → lastn(n, x) = lastn(n -1, x.butlast) + x.last subst: [x, n] → [lastn(n, x), m0 + 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: lastn(n, x), m0 + 1