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Proof step 3 from proof firsteq-01 (previous step)

y + z = a ' + x, a ∈ y ∨ a ∈ z ⊦ y = [], a = y.head ∧ x = y.tail + z

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] → [y, a0, y0]

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%e-01: ⊦ a ' + x ≠ [];%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a0, y0