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Proof step 4 from proof cyclic-def-shortest (previous step)

¬ ¬ validrefs(str, r ' + refs2, st[r -- _type] .type, st), validrefs(str, (r ' + refs2) + r1 ', st[r -- _type] .type, st), ∀ st, r, str, refs. # refs < # (refs2 + r1 ') → ( validrefs(str, r ' + refs, st[r -- _type] .type, st) → (∃ refs0, r0. ¬ dups(refs0) ∧ r0 ∈ refs0 ∧ refs0 .first = r ∧ validrefs(str, refs0 + r0 ', st[refs0 .first -- _type] .type, st)) ∨ ¬ r ∈ refs ∧ ¬ dups(refs)) ⊦ ∃ refs0, r0. ¬ dups(refs0) ∧ r0 ∈ refs0 ∧ refs0 .first = r ∧ validrefs(str, refs0 + r0 ', st[refs0 .first -- _type] .type, st), ¬ r ∈ refs2 ∧ ¬ dups(refs2) ∧ (r1 = r ∨ r1 ∈ refs2), ¬ r ∈ refs2 ∧ r ≠ r1 ∧ ¬ dups(refs2 + r1 '), ¬ r ∈ refs2 ∧ ¬ dups(refs2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
validrefs(str, (r ' + refs2) + r1 ', st[r -- _type] .type, st), validrefs(str, r ' + refs2, st[r -- _type] .type, st), ∀ st, r, str, refs. # refs < # (refs2 + r1 ') → ( validrefs(str, r ' + refs, st[r -- _type] .type, st) → (∃ refs0, r0. ¬ dups(refs0) ∧ r0 ∈ refs0 ∧ refs0 .first = r ∧ validrefs(str, refs0 + r0 ', st[refs0 .first -- _type] .type, st)) ∨ ¬ r ∈ refs ∧ ¬ dups(refs)) ⊦ ¬ r ∈ refs2 ∧ ¬ dups(refs2), ¬ r ∈ refs2 ∧ r ≠ r1 ∧ ¬ dups(refs2 + r1 '), ¬ r ∈ refs2 ∧ ¬ dups(refs2) ∧ (r1 = r ∨ r1 ∈ refs2), ∃ refs0, r0. ¬ dups(refs0) ∧ r0 ∈ refs0 ∧ refs0 .first = r ∧ validrefs(str, refs0 + r0 ', st[refs0 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]