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Proof step 120 from proof junkrefs-union-disjoint (conclusion)

⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + refs0, ty, st1 ∪ st) → junkrefs(str, r ' + refs0, ty, st1)

User applied rule structural induction

And obtained the following new goals

Proof step 19:
⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + [], ty, st1 ∪ st) → junkrefs(str, r ' + [], ty, st1)
Proof step 119:
∀ st1, st, r, str, ty. st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + acc, ty, st1 ∪ st) → junkrefs(str, r ' + acc, ty, st1) ⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + (r0 + acc), ty, st1 ∪ st) → junkrefs(str, r ' + (r0 + acc), ty, st1)

Rule argument

[VARWITHVARSEQSARG : refs0, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : references generated by [], +;])]]