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Proof step 4 from proof firstn-append-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
# x0 = n0 + # z ∧ # z ≤ # x0, a0 ' + x0 = firstn(n0, a0 ' + x0) + a ' + z, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z, a0 ∈ firstn(n0, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ a0 ' + x0 = firstn(# x0 + 1 - # z, a0 ' + x0) + z

Rule argument

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

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%len-03: ⊦ a ' + x = y + b ' + z → ¬ # x < # z; from specification list%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, # z, n