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Proof step 4 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 + 1, x), n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ≠ 0 → lastn(n, x) = lastn(n -1, x.butlast) + x.last
Proof step 6:
m0 + 1 ≠ 0, m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0 + 1, lastn(n, x)) = lastn((m0 + 1)-1, x.butlast) + x.last, n = 0

Rule argument

spec: list-del inst: name: lastn-rec seq: ⊦ n ≠ 0 → lastn(n, x) = lastn(n -1, x.butlast) + x.last subst: [x, n] → [x, m0 + 1] flags: rewrite, current

Additional information

Used simplifier rules: z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: x, m0 + 1