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Proof step 2 from proof c-02 (previous step)

a ' + x = y + z, a ∈ y ∨ a ∈ z ⊦ y = [], a = y.first ∧ x = y.rest + z

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] → [y, a0, y0] flags:

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;%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: y, a0, y0