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Proof step 14 from proof add-nput (conclusion)

⊦ ¬ sk ∈ fis ∧ sk ≠ _type → (addobj(r, class1, fis, st)[r -- sk, val]) = addobj(r, class1, fis, st[r -- sk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ sk = _type, (addobj(r, class1, fis, st)[r -- sk, val]) = addobj(r, class1, fis, st[r -- sk, val]), sk ∈ fis

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]