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Proof step 4 from proof rev-rest (previous step)

rev(x) + y = a ' + y0, a ∈ x ∨ a ∈ y ⊦ x = [], y0 = rev(x).rest + y

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
rev(x) = a0 + y1 ∧ rev(x) ≠ [], rev(x) + y = a ' + y0, a ∈ x ∨ a ∈ y ⊦ x = [], y0 = y1 + y

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] → [rev(x), a0, y1] flags:

Additional information

Used simplifier rules: rev-empty: ⊦ rev(x) = [] ↔ x = [];%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: rev(x), a0, y1