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Proof step 2 from proof ins-less-last (previous step)

⊦ ins≤(a, x + a0 ') = ins≤(a, x) + a0 ', a0 < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ins≤(a, x + a0 ') = ins≤(a, x) + a0 ', a0 < a

Rule argument

Additional information