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

a0 ' + y = firstn(m, a0 ' + y) + a ' + z, # y = m + # z, # z ≤ # y + 1, ∀ z0, a1. a0 ' + y = firstn(# y - # z0, a0 ' + y) + a1 ' + z0 → a0 ' + y = firstn(# y + 1 - # z0, a0 ' + y) + z0, a0 ∈ firstn(m, a0 ' + y) ∨ a0 = a ∨ a0 ∈ z ⊦ y = firstn(m, y) + z, a0 ' + y = firstn(# y - # z, a0 ' + y) + a ' + z

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%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: # y, # z, n