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Proof step 107 from proof validrefs-subset-equiv (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 106:
st ⊆ st0, validref(str, r, st[r -- _type] .type, st) ⊦ validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]