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Proof step 64 from proof validrefs-put-notin (conclusion)

⊦ ¬ r ∈ refs → (validrefs(str, refs, ty, st[r -- sk, val]) ↔ validrefs(str, refs, ty, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 63:
⊦ validrefs(str, refs, ty, st[r -- sk, val]) ↔ validrefs(str, refs, ty, st), r ∈ refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]