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Proof step 34 from proof validrefkeynotnull-subset (conclusion)

⊦ validrefkeynotnull(str, rk, ty, st) ∧ st ⊆ st1 → validrefkeynotnull(str, rk, ty, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
st ⊆ st1, validrefkeynotnull(str, rk, ty, st) ⊦ validrefkeynotnull(str, rk, ty, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]