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Proof step 11 from proof disjoint-nopathex (previous step)

∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, st[r -- _type] .type, st), path∃(str, r0, r1, st), ∀ refs, refs0. refs = [] ∨ refs0 = [] ∨ refs .first ≠ r ∧ refs .first ≠ r0 ∨ refs0 .first ≠ r ∧ refs0 .first ≠ r0 ∨ refs .first = refs0 .first ∨ refs .last ≠ refs0 .last ∨ refs .last = jvmref ∨ ¬ validrefs(str, refs, st[refs .first -- _type] .type, st) ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ⊦ r = r0, r1 = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
validrefs(str, refs, st[refs .first -- _type] .type, st), path∃(str, r0, refs .last, st), ∀ refs1, refs0. refs1 = [] ∨ refs0 = [] ∨ refs1 .first ≠ refs .first ∧ refs1 .first ≠ r0 ∨ refs0 .first ≠ refs .first ∧ 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) ⊦ refs .first = r0, refs .last = jvmref, refs = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]