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Proof step 24 from proof add-dupput (previous step)

sk ∈ fis ⊦ (r0 -- sk0 ∈ addobj(r, class1, fis + (sk × val) ' + fis0, st) ↔ r0 -- sk0 ∈ addobj(r, class1, fis + fis0, st)) ∧ (addobj(r, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = (addobj(r, class1, fis + fis0, st)[r0 -- sk0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 21:
¬ r ≠ r0, sk ∈ fis ⊦ (r0 -- sk0 ∈ addobj(r, class1, fis + (sk × val) ' + fis0, st) ↔ r0 -- sk0 ∈ addobj(r, class1, fis + fis0, st)) ∧ (addobj(r, class1, fis + (sk × val) ' + fis0, st)[r0 -- sk0]) = (addobj(r, class1, fis + fis0, st)[r0 -- sk0])
Proof step 22:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 23:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[crewrite-arg: add-ref , F , [sk, r0, st, fis, class1, r] → [sk0, r0, st, fis + (sk × val) ' + fis0, class1, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk0, r0, st, fis + (sk × val) ' + fis0, class1, r)]]