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Proof step 2 from proof eq (previous step)

ins≤(a, x) = ins≤(a, y) ⊦ x = y

User applied rule structural induction

And obtained the following new goals

Proof step 3:
ins≤(a, []) = ins≤(a, y) ⊦ [] = y
Proof step 7:
ins≤(a, a0 + x0) = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 + x0 = y

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;