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Proof step 9 from proof disjoint-def-pathex (previous step)

∀ refs0, refs1. refs0 = [] ∨ refs1 = [] ∨ ¬ refs0 .first ∈ refs ∨ ¬ refs1 .first ∈ refs ∨ refs0 .first = refs1 .first ∨ refs0 .last ≠ refs1 .last ∨ refs0 .last = jvmref ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ⊦ ∀ r, r0, r1. ¬ r ∈ refs ∨ ¬ r0 ∈ refs ∨ r = r0 ∨ r1 = jvmref ∨ ¬ path∃(str, r, r1, st) ∨ ¬ path∃(str, r0, r1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
path∃(str, r0, r, st), path∃(str, r1, r, st), r0 ∈ refs, r1 ∈ refs, ∀ refs0, refs1. refs0 = [] ∨ refs1 = [] ∨ ¬ refs0 .first ∈ refs ∨ ¬ refs1 .first ∈ refs ∨ refs0 .first = refs1 .first ∨ refs0 .last ≠ refs1 .last ∨ refs0 .last = jvmref ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ⊦ r = jvmref, r1 = r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]