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Proof step 21 from proof disjoint-nojunk-newref-subset-01 (previous step)

(r1 ' + refs1) .last = (r ' + refs0) .last, validrefs(str, r ' + refs0, st1[r -- _type] .type, st1), validrefs(str, r1 ' + refs1, st1[r1 -- _type] .type, st), newref(r, st), st ⊆ st1, path∃(str, r, (r ' + refs0) .last, st1) → newref((r ' + refs0) .last, st) ⊦ path∃(str, r, (r ' + refs0) .last, st1), (r ' + refs0) .last = jvmref, r = jvmref, r1 = r, junkref(str, r1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
(r1 ' + refs1) .last = (r ' + refs0) .last, validrefs(str, r ' + refs0, st1[r -- _type] .type, st1), validrefs(str, r1 ' + refs1, st1[r1 -- _type] .type, st), newref(r, st), st ⊆ st1 ⊦ (r ' + refs0) .last = jvmref, r = jvmref, r1 = r, path∃(str, r, (r ' + refs0) .last, st1), junkref(str, r1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]