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Proof step 2 from proof mk-first (previous step)

⊦ (mklist(a, n) + x).first = a, n = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [mklist(a, n) + x, a0, y] flags:

Additional information

Used simplifier rules: mk-empty: ⊦ mklist(a, m) = [] ↔ m = 0;%mk-elem: ⊦ a ∈ mklist(b, n) → a = b;%mk-empty: ⊦ mklist(a, m) = [] ↔ m = 0;%c: ⊦ a + x = a ' + x; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-append0-not-left: ⊦ a ' + x = x0 + x1 → (¬ a ∈ x0 ↔ x0 = []); from specification list%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: mklist(a, n) + x, a0, y