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Proof step 14 from proof validrefs-nodups (previous step)

∃ refs3. validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref ∧ (r ' + refs) .first = refs3 .first ∧ ¬ dups(refs3) ∧ refs3 ⊆ r ' + refs, refs .last = jvmref, validrefs(str, r ' + refs0 + r ' + refs, ty, st), validrefs(str, r ' + refs, ty, st), ∀ st, ty, str, refs1. # refs1 < # (r + (refs0 + (r + refs))) → ( validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = jvmref → (∃ refs3. validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref ∧ refs1 .first = refs3 .first ∧ ¬ dups(refs3) ∧ refs3 ⊆ refs1)) ⊦ refs = [], ∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = jvmref ∧ r = refs1 .first ∧ ¬ dups(refs1) ∧ refs1 ⊆ refs1 .first ' + refs0 + refs1 .first ' + refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
refs1 .last = jvmref, refs .last = jvmref, validrefs(str, refs1 .first ' + refs, ty, st), validrefs(str, refs1 .first ' + refs0 + refs1 .first ' + refs, ty, st), refs1 ⊆ refs1 .first ' + refs, validrefs(str, refs1, ty, st), ∀ st, ty, str, refs2. # refs2 < # (refs1 .first + (refs0 + (refs1 .first + 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 ∧ refs1 .first = refs2 .first ∧ ¬ dups(refs2) ∧ refs2 ⊆ refs2 .first ' + refs0 + refs2 .first ' + refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ (r ' + refs) .first = r])]]