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Proof step 5 from proof add-length (previous step)

⊦ (addobj(r0, class, fis, st)[r -- _length]) = st[r -- _length], _length ∈ fis

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 4:
¬ r0 ≠ r ⊦ (addobj(r0, class, fis, st)[r -- _length]) = st[r -- _length], _length ∈ fis

Rule argument

[crewrite-arg: add-ref , F , [sk, r0, st, fis, class1, r] → [_length, r, st, fis, class, r0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (_length, r, st, fis, class, r0)]]