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Proof step 143 from proof validrefs-put-notin-03 (previous step)

refs0 = r0 + refs1, is_valid_ref(str, r, st) ↔ is_valid_ref(str, r, st[r -- sk, val]), list2set(refs) ∩ list2set(refs0) = ∅ ⊦ val .refval = r0, r0 = st[r -- sk] .refval, refs0 = [], sk = _type, sk = _length, validrefs(str, refs + r ' + refs0, ty, st[r -- sk, val]) ↔ validrefs(str, refs + r ' + refs0, ty, st), r ∈ refs0, r ∈ refs, dups(refs0), dups(refs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list-dup, references, dups-cons, ⊦ dups(r ' + refs) ↔ dups(refs) ∨ r ∈ refs], [CSIMPNAMEDSEQ : list, references, in-06, ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPNAMEDSEQ : list, references, e-01, ⊦ r ' + refs ≠ []], [CSIMPNAMEDSEQ : set-union, references, int-c, ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPNAMEDSEQ : set-union, references, int-a, ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1], [CSIMPNAMEDSEQ : setlist, references, in, ⊦ r ∈ list2set(refs) ↔ r ∈ refs], [CSIMPNAMEDSEQ : set-union, references, int-ins, ⊦ rset ∩ (rset0 ++ r) = ∅ ↔ rset ∩ rset0 = ∅ ∧ ¬ r ∈ rset], [CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : setlist, references, list2set-rec, ⊦ list2set(r ' + refs) = list2set(refs) ++ r], [CSIMPNAMEDSEQ : list, references, c, ⊦ r + refs = r ' + refs])]]