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Proof step 26 from proof validrefs-nodups (previous step)

refs = [], (r0 ' + refs) .last = jvmref, validrefs(str, r ' + refs1 + r0 ' + refs0 + r0 ' + refs, ty, st), ∀ st, ty, str, refs3. # refs3 < # (r + (refs1 + (r0 + (refs0 + (r0 + refs))))) → ( validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = jvmref ∧ refs3 .first = refs1 .first ∧ ¬ dups(refs1) ∧ refs1 ⊆ refs3)) ⊦ r = r0, r ∈ refs, r ∈ refs0, r ∈ refs1, ∃ refs2. validrefs(str, refs2, ty, st) ∧ refs2 ≠ [] ∧ refs2 .last = jvmref ∧ r = refs2 .first ∧ ¬ dups(refs2) ∧ refs2 ⊆ refs2 .first ' + refs1 + r0 ' + refs0 + r0 ' + refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
validrefs(str, r ' + refs1 + jvmref ' + refs0 + jvmref ', ty, st), ∀ st, ty, str, refs3. # refs3 < # (r + (refs1 + (jvmref + (refs0 + (jvmref + []))))) → ( validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] ∧ refs3 .last = jvmref → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = jvmref ∧ refs3 .first = refs1 .first ∧ ¬ dups(refs1) ∧ refs1 ⊆ refs3)) ⊦ r = jvmref, r ∈ refs1, r ∈ refs0, ∃ refs. validrefs(str, refs, ty, st) ∧ refs ≠ [] ∧ refs .last = jvmref ∧ r = refs .first ∧ ¬ dups(refs) ∧ refs ⊆ refs .first ' + refs1 + jvmref ' + refs0 + jvmref '

Rule argument

[emptyarg]

Additional information

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