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Proof step 11 from proof add-gins (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ addobj(r, class1, (sk × val) ' + fis, st) = (addobj(r, class1, fis, st)[r -- sk, val])
Proof step 10:
∀ r, class1, st, sk, val. (addobj(r, class1, fis0, st)[r -- sk, val]) = addobj(r, class1, fis0[sk, val], st) ⊦ sk = sk0, (addobj(r, class1, fis0, st)[r -- sk0, val0][r -- sk, val]) = (addobj(r, class1, fis0[sk, val], st)[r -- sk0, val0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]