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Proof step 8 from proof disjoint-nopathex (previous step)

validrefs(str, r ' + refs2, st[r -- _type] .type, st), path∃(str, r0, (r ' + refs2) .last, st), ∀ refs1, refs0. refs1 = [] ∨ refs0 = [] ∨ refs1 .first ≠ r ∧ refs1 .first ≠ r0 ∨ refs0 .first ≠ r ∧ refs0 .first ≠ r0 ∨ refs1 .first = refs0 .first ∨ refs1 .last ≠ refs0 .last ∨ refs1 .last = jvmref ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ⊦ r = r0, (r ' + refs2) .last = jvmref

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st))
Proof step 7:
validrefs(str, r ' + refs2, st[r -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r0 ∧ refs .last = (r ' + refs2) .last ∧ validrefs(str, refs, st[r0 -- _type] .type, st), ∀ refs1, refs0. refs1 = [] ∨ refs0 = [] ∨ refs1 .first ≠ r ∧ refs1 .first ≠ r0 ∨ refs0 .first ≠ r ∧ refs0 .first ≠ r0 ∨ refs1 .first = refs0 .first ∨ refs1 .last ≠ refs0 .last ∨ refs1 .last = jvmref ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ⊦ r = r0, (r ' + refs2) .last = jvmref

Rule argument

[speclemmaarg: , path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st)) ,[st, r0, r, str] → [st, (r ' + refs2) .last, r0, str] ,F ,cyclic , ,pathex-def-simple ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, (r ' + refs2) .last, r0, str)]]