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Proof step 2 from proof validrefkeynotnull-union-same-02 (previous step)

validrefkeynotnull(str, rk, ty, st1) ⊦ (st ∪ st1)[rk] = st1[rk]

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ validrefkeynotnull(str, rk, ty, st1) → st1[rk] = (st ∪ st1)[rk]

Rule argument

[LEMMAARG : validrefkeynotnull-union-same-01, [str, rk, ty, st1, st] → [str, rk, ty, st1, st]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (str, rk, ty, st1, st)]]