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Proof step 63 from proof junkrefs-subset (previous step)

∀ str, ty, st1, st. junkrefs(str, acc, ty, st1) ∧ st ⊆ st1 → junkrefs(str, acc, ty, st), junkrefs(str, r + acc, ty, st1), st ⊆ st1 ⊦ junkrefs(str, r + acc, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
st ⊆ st1, junkrefs(str, r ' + acc, ty, st1), ∀ str, ty, st1, st. junkrefs(str, acc, ty, st1) ∧ st ⊆ st1 → junkrefs(str, acc, ty, st) ⊦ junkrefs(str, r ' + acc, ty, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r + refs = r ' + refs])]]