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Proof step 10 from proof cyclic-put-nopathex (previous step)

¬ (¬ path∃(str, r, r0, st[r0 -- sk, val]) ∧ r0 ∈ refs1 + r1 '), validrefs(str, (r ' + refs1) + r1 ', st[r -- _type] .type, st[r0 -- sk, val]) ⊦ sk = _type, r0 = r, validrefs(str, (r ' + refs1) + r1 ', st[r -- _type] .type, st), path∃(str, r, r0, st), junkref(str, r, st), dups(refs1), r ∈ refs1, r1 ≠ r ∧ ¬ r1 ∈ refs1, r0 ≠ r1 ∧ ¬ r0 ∈ refs1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
validrefs(str, (r ' + refs1) + r1 ', st[r -- _type] .type, st[r0 -- sk, val]) ⊦ sk = _type, r0 = r, validrefs(str, (r ' + refs1) + r1 ', st[r -- _type] .type, st), path∃(str, r, r0, st), junkref(str, r, st), dups(refs1), r ∈ refs1, r1 ≠ r ∧ ¬ r1 ∈ refs1, r0 ≠ r1 ∧ ¬ r0 ∈ refs1, ¬ path∃(str, r, r0, st[r0 -- sk, val]) ∧ (r0 ∈ refs1 ∨ r0 = r1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)])]]