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Proof step 11 from proof junkref-union-disjoint (previous step)

¬ ¬ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, junkrefs(str, r ' + refs0, st[r -- _type] .type, st1 ∪ st), r ∈ st ⊦ junkrefs(str, r ' + refs0, st1[r -- _type] .type, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPSEQ : ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1], [CSIMPSEQ : ⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]])]]