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Proof step 1 from proof del1-01-02 (conclusion)

⊦ a ≠ b ∧ ordered≤(x) → ins≤(a, x -1l b) = ins≤(a, x) -1l b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
ordered≤(x) ⊦ ins≤(a, x -1l b) = ins≤(a, x) -1l b, a = b

Rule argument

Additional information