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Proof step 54 from proof disjoint-nojunk-newref-subset-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 53:
(r1 ' + refs0) .last = (r ' + refs1) .last, r1 ∈ st, st ⊆ st1, newref(r, st), validrefs(str, r ' + refs1, st1[r -- _type] .type, st1), validrefs(str, r1 ' + refs0, st[r1 -- _type] .type, st1), r1 -- _type ∈ st ⊦ r = r1, r = jvmref, (r ' + refs1) .last = jvmref, path∃(str, r, (r ' + refs1) .last, st1), junkref(str, r1, st), ¬ path∃(str, r, refs1 .last, st1) ∧ (r ' + refs1) .last ∈ refs1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ∈ refs → (r0 ' + refs) .last = refs .last])]]