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Proof step 2 from proof abs-leb-eq-mapped (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 3:
abs(leb) = mapped(buf), leb = mapped(buf0, m) ⊦ ∃ buf, n. leb = mapped(buf, n)

Rule argument

var: leb

Additional information

Used simplifier rules: abs-leb-unmapped: ⊦ abs(unmapped) = unmapped;%abs-leb-erased: ⊦ abs(erased) = unmapped;%leb freely generated by unmapped, erased, mapped;;% ⊦ unmapped ≠ mapped(buf); (s)