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Proof step 17 from proof firstn-last-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + firstn(# x0 - # z, x0).last ' + z, a0 ∈ firstn(n1, a0 ' + x0), # z ≤ # x0 ⊦ firstn(n1, x0).last = firstn(# x0 - # z, x0).last, n1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ x → a ∈ x + y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list