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Proof step 3 from proof app-05 (previous step)

a ∈ x ⊦ pos(a, x + y) = pos(a, x)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
a ∈ a0 + x0, ∀ a, y. a ∈ x0 → pos(a, x0 + y) = pos(a, x0) ⊦ pos(a, (a0 + x0) + y) = pos(a, a0 + x0)

Rule argument

term: x

Additional information

Used simplifier rules: pos-e: ⊦ pos(a, []) = 0;%list freely generated by [], +;;% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ ¬ a ∈ []; (s)