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Proof step 16 from proof put-firstn-restn (previous step)

a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ z0 = [], z0.rest = z

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
z0 = a + y1, a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ z0 = [], y1 = z

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] → [z0, a, y1] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: z0, a, y1