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Proof step 13 from proof junkref-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r ∈ st1, junkrefs(str, r ' + refs0, (st1 ∪ st)[r -- _type] .type, st1 ∪ st) ⊦ junkrefs(str, r ' + refs0, st1[r -- _type] .type, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]
Proof step 11:
¬ ¬ r -- _type ∈ st, st1 .refs ∩ st .refs = ∅, r ∈ st1, junkrefs(str, r ' + refs0, (st1 ∪ st)[r -- _type] .type, st1 ∪ st) ⊦ junkrefs(str, r ' + refs0, st1[r -- _type] .type, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)
Proof step 12:
⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ junkrefs(str, r ' + refs0, ty, st1 ∪ st) → junkrefs(str, r ' + refs0, ty, st1)

Rule argument

[speclemmaarg: ,¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk] ,[rk, st0, st] → [r -- _type, st, st1] ,F ,storefunsx , ,union-at-left ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1], [CSIMPSEQ : ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (r -- _type, st, st1)]]