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Proof step 13 from proof firstn-last-01 (previous step)

# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, a0 ∈ firstn(n1, a0 ' + x0), # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a ⊦ firstn(n1, x0).last = a, firstn(n1, x0) = [], n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, a0 ∈ firstn(n1, a0 ' + x0), # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a ⊦ firstn(n1, x0).last = a, n1 = 0

Rule argument

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;