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Proof step 18 from proof junkref-union-disjoint (previous step)

refs = r + refs0, st1 .refs ∩ st .refs = ∅, junkrefs(str, refs, (st1 ∪ st)[r -- _type] .type, st1 ∪ st), r ∈ st1 ⊦ refs = [], junkref(str, r, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
st1 .refs ∩ st .refs = ∅, r ∈ st1, junkrefs(str, r ' + refs0, (st1 ∪ st)[r -- _type] .type, st1 ∪ st) ⊦ junkref(str, r, st1)

Rule argument

[emptyarg]

Additional information

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