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Proof step 3 from proof last-sizeone (previous step)

# x = 1 ⊦ (y + x).last = x.head

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x, a, y0]

Additional information

Used simplifier rules: lapp-01: ⊦ (x + a ' + y).last = (a ' + y).last;%last-one: ⊦ a '.last = a;% ⊦ a + x = a ' + x; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a ' + x ≠ []; (s)% ⊦ x + [] = x; (s)% ⊦ # [] = 0; (s)%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a, y0