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Proof step 12 from proof disjoint-def-pathex (previous step)

(r2 ' + refs2) .last = (r3 ' + refs1) .last, validrefs(str, r3 ' + refs1, st[r3 -- _type] .type, st), validrefs(str, r2 ' + refs2, st[r2 -- _type] .type, st), r3 ∈ refs, r2 ∈ refs ⊦ (r3 ' + refs1) .last = jvmref, r2 = r3, ∃ refs. refs ≠ [] ∧ refs .first = r2 ∧ refs .last = (r3 ' + refs1) .last ∧ validrefs(str, refs, st[r2 -- _type] .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(r2 ' + refs2) .last = (r3 ' + refs1) .last, validrefs(str, r3 ' + refs1, st[r3 -- _type] .type, st), validrefs(str, r2 ' + refs2, st[r2 -- _type] .type, st), r3 ∈ refs, r2 ∈ refs ⊦ (r3 ' + refs1) .last = jvmref, r2 = r3, ∃ refs0. refs0 ≠ [] ∧ refs0 .first = r2 ∧ refs0 .last = (r3 ' + refs1) .last ∧ validrefs(str, refs0, st[refs0 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]