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Proof step 15 from proof firstn-last-01 (previous step)

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

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, 1 + # z, n] flags:

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;%firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%c: ⊦ a + x = a ' + x; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%len-06: ⊦ # y = # y0 → (x + y = z + y0 ↔ x = z ∧ y = y0); from specification list%c-05: ⊦ a ' = x + b ' ↔ a = b ∧ x = []; from specification list%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, 1 + # z, n