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Proof step 3 from proof validrefkeynotnull-union-same (conclusion)

⊦ validrefkeynotnull(str, r -- fs ', ty, st1) → st1[r -- fs '] = (st ∪ st1)[r -- fs ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
validrefkeynotnull(str, r -- fs ', ty, st1) ⊦ st1[r -- fs '] = (st ∪ st1)[r -- fs ']

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]