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Proof step 8 from proof firstn-last-01 (previous step)

# x0 = n0 + # z, # x0 = n1 + # 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 ⊦ (a0 ' + firstn(n1, x0)).last = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a, a0 ∈ firstn(n1, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ (a0 ' + firstn(n1, x0)).last = a

Rule argument

Additional information

Used simplifier rules: eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic