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Proof step 6 from proof validrefs-subset-equiv-01 (conclusion)

⊦ ¬ junkref(str, r, 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 5:
st ⊆ st0 ⊦ validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st), junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]