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Proof step 12 from proof pathex-def-nodups (previous step)

validrefs(str, r ' + refs1, st[r -- _type] .type, st), ∃ refs0, refs2. refs1 = refs0 + (r + refs2) ⊦ refs1 = [], r = refs1 .last, dups(refs1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = refs1 .last ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
list2set(refs0) ∩ list2set(refs) = ∅, validrefs(str, r ' + refs0 + r ' + refs, st[r -- _type] .type, st) ⊦ r = (r ' + refs) .last, r ∈ refs, r ∈ refs0, dups(refs), dups(refs0), ∃ refs1. refs1 ≠ [] ∧ refs1 .first = r ∧ refs1 .last = (refs1 .first ' + refs) .last ∧ ¬ dups(refs1) ∧ validrefs(str, refs1, st[refs1 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ dups(refs) ∧ ¬ dups(refs0) → ( dups(refs + refs0) ↔ list2set(refs) ∩ list2set(refs0) ≠ ∅)], [CSIMPFORWARD : ⊦ ¬ dups(refs + refs0) → ¬ dups(refs) ∧ ¬ dups(refs0)], [CSIMPSEQ : ⊦ dups(refs + r ' + refs0) ↔ r ∈ refs ∨ r ∈ refs0 ∨ dups(refs + refs0)], [CSIMPSEQ : ⊦ (refs + r ' + refs0) .last = (r ' + refs0) .last], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ refs + r ' + refs0 ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]