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Proof step 12 from proof validrefs-nodups (previous step)

refs1 = r + refs4, refs1 .last = jvmref, refs .last = jvmref, validrefs(str, r ' + refs, ty, st), validrefs(str, r ' + refs0 + r ' + refs, ty, st), refs1 ⊆ r ' + refs, validrefs(str, refs1, 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)) ⊦ refs1 = [], refs = [], dups(refs1), ∃ refs2. validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref ∧ r = refs2 .first ∧ ¬ dups(refs2) ∧ refs2 ⊆ refs2 .first ' + refs0 + refs2 .first ' + refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
refs .last = jvmref, (r ' + 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)) ⊦ refs = [], dups(refs4), r ∈ refs4, ∃ refs2. validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref ∧ r = refs2 .first ∧ ¬ dups(refs2) ∧ refs2 ⊆ refs2 .first ' + refs0 + refs2 .first ' + refs

Rule argument

[emptyarg]

Additional information

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