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Proof step 7 from proof add-nput-02 (previous step)

∀ r, class1, sk, val, st. addobj(r, class1, fis0 + (sk × val) ', st) = (addobj(r, class1, fis0, st)[r -- sk, val]) ∨ sk ∈ fis0 ⊦ (addobj(r, class1, fis0 + (sk × val) ', st)[r -- sk0, val0]) = (addobj(r, class1, fis0, st)[r -- sk0, val0][r -- sk, val]), sk = sk0, sk ∈ fis0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ rk ≠ rk0 → (st[rk0, val][rk, val0]) = (st[rk, val0][rk0, val])
Proof step 6:
r -- sk ≠ r -- sk0, ∀ r, class1, sk, val, st. addobj(r, class1, fis0 + (sk × val) ', st) = (addobj(r, class1, fis0, st)[r -- sk, val]) ∨ sk ∈ fis0 ⊦ (addobj(r, class1, fis0 + (sk × val) ', st)[r -- sk0, val0]) = (addobj(r, class1, fis0, st)[r -- sk, val][r -- sk0, val0]), sk = sk0, sk ∈ fis0

Rule argument

[speclemmaarg: ,rk ≠ rk0 → (st[rk0, val][rk, val0]) = (st[rk, val0][rk0, val]) ,[val0, rk, val, rk0, st] → [val, r -- sk, val0, r -- sk0, addobj(r, class1, fis0, st)] ,F ,store ,javastore ,rew ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0])], [EXTRATERMS : (val, r -- sk, val0, r -- sk0, addobj(r, class1, fis0, st))]]