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Proof step 10 from proof lastn-restn (previous step)

m + m0 ≤ # x0, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(m0, x0)) = lastn(m, a ' + x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ n < # x → lastn(n, x) = lastn(n, x.tail)
Proof step 12:
m < #(a ' + x0), m + m0 ≤ # x0, ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(m0, x0)) = lastn(m, (a ' + x0).tail)

Rule argument

spec: <toplevel> name: lastN-rec-02 seq: ⊦ n < # x → lastn(n, x) = lastn(n, x.tail) subst: [x, n] → [a ' + x0, 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 + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s) Used terms: a ' + x0, m