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Proof step 2 from proof add-gins-01 (conclusion)

sk0 ≠ sk ⊦ addobj(r, class1, ((sk0 × val0) ' + fis)[sk, val], st) = addobj(r, class1, (sk0 × val0) ' + fis[sk, val], st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ addobj(r, class1, (sk × val) ' + fis, st) = (addobj(r, class1, fis, st)[r -- sk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ sk ≠ sk0 → ((sk0 × val) ' + fis)[sk, val0] = (sk0 × val) ' + fis[sk, val0]])]]