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Proof step 2 from proof id-idx-len-eq-1 (previous step)

n = # lpt1, id-idx(lpt0, lpt1) ⊦ n = # lpt0 ↔ true

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: id-idx-len-eq: ⊦ id-idx(lpt0, lpt1) → (# lpt0 = # lpt1 ↔ true);