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Proof step 5 from proof firstn-last-01 (previous step)

a0 ' + x0 = firstn(# x0 - # z, a0 ' + x0) + a ' + z, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a, a0 ∈ firstn(# x0 - # z, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ firstn(1 + # x0 - # z, a0 ' + x0).last = a

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
# x0 = n0 + # z, a0 ' + x0 = firstn(n0, a0 ' + x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a, a0 ∈ firstn(n0, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ firstn(1 + # x0 - # z, a0 ' + x0).last = a

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# x0, # z, n]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = y + b ' + z → ¬ # x < # z; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, # z, n