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Proof step 65 from proof disjoint-nojunk-newref-subset-01 (previous step)

(r2 ' + refs1) .last = (r3 ' + refs0) .last, st ⊆ st1, newref(r, st), validrefs(str, r2 ' + refs1, st1[r2 -- _type] .type, st1), validrefs(str, r3 ' + refs0, st1[r3 -- _type] .type, st1), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r2 = r3, r = jvmref, (r3 ' + refs0) .last = jvmref, junkref(str, r1, st), r2 ≠ r1 ∧ r2 ≠ r, r3 ≠ r1 ∧ r3 ≠ r

User applied rule case distinction

And obtained the following new goals

Proof step 27:
(r2 ' + refs1) .last = (r3 ' + refs0) .last, st ⊆ st1, newref(r, st), validrefs(str, r2 ' + refs1, st1[r2 -- _type] .type, st1), validrefs(str, r3 ' + refs0, st1[r3 -- _type] .type, st1), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r2 ≠ r1, r2 = r3, r = jvmref, (r3 ' + refs0) .last = jvmref, junkref(str, r1, st), r3 ≠ r1 ∧ r3 ≠ r
Proof step 64:
(r2 ' + refs1) .last = (r3 ' + refs0) .last, st ⊆ st1, newref(r, st), validrefs(str, r2 ' + refs1, st1[r2 -- _type] .type, st1), validrefs(str, r3 ' + refs0, st1[r3 -- _type] .type, st1), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r2 ≠ r, r2 = r3, r = jvmref, (r3 ' + refs0) .last = jvmref, junkref(str, r1, st), r3 ≠ r1 ∧ r3 ≠ r

Rule argument

[CASEDARG : pos: right 5, ()]

Additional information

[[CONCRETESIMPRULES : ()]]