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Proof step 119 from proof junkrefs-union-disjoint (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 118:
st1 .refs ∩ st .refs = ∅, r ∈ st1, junkrefs(str, r ' + r0 ' + acc, ty, st1 ∪ st), ∀ st1, st, r, str, ty. st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + acc, ty, st1 ∪ st) → junkrefs(str, r ' + acc, ty, st1) ⊦ junkrefs(str, r ' + r0 ' + acc, ty, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]