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Proof step 8 from proof validrefs-nodups (previous step)

refs .last = jvmref, refs4 .last = jvmref, validrefs(str, r ' + refs, ty, st), validrefs(str, r ' + refs0 + r ' + refs, ty, st), refs4 ⊆ r ' + refs, validrefs(str, r ' + refs4, ty, st), ∀ st, ty, str, refs2. # refs2 < # (r + (refs0 + (r + refs))) → ( validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref → (∃ refs3. validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref ∧ refs2 .first = refs3 .first ∧ ¬ dups(refs3) ∧ refs3 ⊆ refs2)) ⊦ refs4 = [], refs = [], dups(refs4), r ∈ refs4, validrefs(str, r ' + refs4, ty, st) ∧ r ' + refs4 ≠ [] ∧ (r ' + refs4) .last = jvmref ∧ r = (r ' + refs4) .first ∧ ¬ dups(r ' + refs4) ∧ r ' + refs4 ⊆ (r ' + refs4) .first ' + refs0 + (r ' + refs4) .first ' + refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

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