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Proof step 4 from proof add-put-02 (conclusion)

sk ≠ sk0 ⊦ (addobj(r, class1, (sk0 × val1) ' + (sk × val) ' + fis, st)[r -- sk, val0]) = addobj(r, class1, (sk0 × val1) ' + (sk × val0) ' + fis, 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])
Proof step 3:
⊦ sk = sk0, (addobj(r, class1, fis, st)[r -- sk0, val1][r -- sk, val0]) = (addobj(r, class1, fis, st)[r -- sk, val0][r -- sk0, val1])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (st[rk, val][rk0, val0][rk, val1] ) = (st[rk0, val0][rk, val1])], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]