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Proof step 80 from proof validrefs-put-notin-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 79:
is_valid_ref(str, r, st) ↔ is_valid_ref(str, r, st[r -- sk, val]) ⊦ r = r0, val .refval = r0, r0 = st[r -- sk] .refval, sk = _type, sk = _length, validrefs(str, r ' + r0 ' + refs1, ty, st[r -- sk, val]) ↔ validrefs(str, r ' + r0 ' + refs1, ty, st), r ∈ refs1, dups(refs1), r0 ∈ refs1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, in-empty, ⊦ ¬ r ∈ []], [CSIMPNAMEDSEQ : list-dup, references, e-01, ⊦ ¬ dups([])], [CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, Nil, ⊦ [] + refs = refs], [CSIMPNAMEDSEQ : set-union, references, int-c, ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPNAMEDSEQ : set-union, references, int-a, ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1], [CSIMPNAMEDSEQ : set-union, references, int-e, ⊦ rset ∩ ∅ = ∅], [CSIMPNAMEDSEQ : setlist, references, list2set-empty, ⊦ list2set([]) = ∅])]]