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Proof step 3 from proof abs-leb-eq-mapped (previous step)

abs(leb) = mapped(buf), leb = mapped(buf0, m) ⊦ ∃ buf, n. leb = mapped(buf, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ∃ buf, n. mapped(buf0, m) = mapped(buf, n)

Rule argument

Additional information

Used simplifier rules: abs-leb-mapped: ⊦ abs(mapped(buf, n)) = mapped(subarray(buf, 0, n));% ⊦ mapped(buf) = mapped(buf0) ↔ buf = buf0; (s)