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Proof step 10 from proof firstn-append-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# x0 = n1 + # z, x0 = firstn(n1 - 1, x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z ⊦ x0 = firstn(n1, x0) + z, n1 = 0

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat