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Proof step 32 from proof validrefkeynotnull-union (conclusion)

⊦ validrefkeynotnull(str, rk, ty, st) ∧ (∃ st0. st1 = st0 ∪ st) → validrefkeynotnull(str, rk, ty, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
validrefkeynotnull(str, rk, ty, st) ⊦ validrefkeynotnull(str, rk, ty, st0 ∪ st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]