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Proof step 4 from proof add-ref (previous step)

⊦ (∀ r, r0, class1, st, sk. r ≠ r0 → (addobj(r, class1, fis0, st)[r0 -- sk]) = st[r0 -- sk]) → r = r0 ∨ (addobj(r, class1, (sk × val) ' + fis0, st)[r0 -- sk0]) = st[r0 -- sk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ addobj(r, class1, (sk × val) ' + fis, st) = (addobj(r, class1, fis, st)[r -- sk, val])
Proof step 3:
∀ r, r0, class1, st, sk. r ≠ r0 → (addobj(r, class1, fis0, st)[r0 -- sk]) = st[r0 -- sk] ⊦ r = r0, (addobj(r, class1, fis0, st)[r0 -- sk0]) = st[r0 -- sk0]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ≠ r0 → (st[r0 -- sk0, val][r -- sk]) = st[r -- sk]], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]