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

# 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

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] → [# x0, # z, n] flags:

Additional information

Used simplifier rules: in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%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%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, # z, n