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Proof step 3 from proof validrefkeynotnull-union-01 (previous step)

validrefkeynotnull(str, rk, ty, st) ∧ (∃ st2. st0 ∪ st = st2 ∪ st) → validrefkeynotnull(str, rk, ty, st0 ∪ st) ⊦ validrefkeynotnull(str, rk, ty, st) → validrefkeynotnull(str, rk, ty, st0 ∪ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
validrefkeynotnull(str, rk, ty, st) ⊦ validrefkeynotnull(str, rk, ty, st0 ∪ st), ∃ st2. st0 ∪ st = st2 ∪ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]