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Proof step 24 from proof disjoint-nojunk-newref-subset-01 (previous step)

(r1 ' + refs1) .last = (r ' + refs0) .last, st ⊆ st1, newref(r, st), validrefs(str, r1 ' + refs1, st1[r1 -- _type] .type, st), validrefs(str, r ' + refs0, st1[r -- _type] .type, st1), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r1 = r, r = jvmref, (r ' + refs0) .last = jvmref, junkref(str, r1, st)

User applied rule all left

And obtained the following new goals

Proof step 23:
(r1 ' + refs1) .last = (r ' + refs0) .last, st ⊆ st1, newref(r, st), validrefs(str, r1 ' + refs1, st1[r1 -- _type] .type, st), validrefs(str, r ' + refs0, st1[r -- _type] .type, st1), path∃(str, r, (r1 ' + refs1) .last, st1) → newref((r1 ' + refs1) .last, st) ∨ (r1 ' + refs1) .last = jvmref ⊦ r1 = r, r = jvmref, (r ' + refs0) .last = jvmref, junkref(str, r1, st)

Rule argument

[EXRARG : pos: left 6, [extquanttermlist: ((r1 ' + refs1) .last) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : ((r1 ' + refs1) .last)]]