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Proof step 39 from proof validrefs-delete-03 (conclusion)

⊦ validrefs(str, refs, ty, st) ∧ ¬ r ∈ refs → validrefs(str, refs, ty, st -- r)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
validrefs(str, refs, ty, st) ⊦ r ∈ refs, validrefs(str, refs, ty, st -- r)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]