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Proof step 106 from proof validrefs-subset-equiv (previous step)

st ⊆ st0, validref(str, r, st[r -- _type] .type, st) ⊦ validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st)

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]