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Proof step 32 from proof disjoint-nojunk-newref-subset-01 (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

[fma-arg: refs0 = []]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ refs .last ∈ refs ↔ refs ≠ []], [CSIMPSEQ : ⊦ refs ≠ [] → (r ' + refs) .last = refs .last], [CSIMPSEQ : ⊦ refs + [] = refs], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' .last = r], [CSIMPSEQ : ⊦ newref(r, st) ∧ r0 ∈ st ∧ r0 ≠ jvmref → r ≠ r0], [CSIMPSEQ : ⊦ r ∈ st → ¬ newref(r, st)])], [EXTRAFMAS : (refs0 = [])]]