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Proof step 3 from proof pos-equal (previous step)

pos(b, x) = pos(a, x), a ∈ x ⊦ a = b

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: x

Additional information

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