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Proof step 50 from proof disjoint-nojunk-newref-subset-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 49:
st ⊆ st1, r1 ∈ st, r1 -- _type ∈ st, validrefs(str, r1 ' + refs0, st[r1 -- _type] .type, st1), validrefs(str, refs0 .last ', st1[refs0 .last -- _type] .type, st1), newref(refs0 .last, st) ⊦ refs0 = [], refs0 .last = jvmref, refs0 .last = r1, junkref(str, r1, st), is_valid_ref(str, refs0 .last, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ newref(r, st) → r ≠ jvmref], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ refs ≠ [] → (r ' + refs) .last = refs .last])]]