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Proof step 9 from proof hasSharing-different-disjoint (previous step)

refs = r3 + refs0, (r0 ' + refs1) .last = refs .last, validrefs(str, refs, st[r3 -- _type] .type, st), path∃(str, r, r2, st), path∃(str, r, r1, st), validrefs(str, r0 ' + refs1, st[r0 -- _type] .type, st) ⊦ refs .last = jvmref, r0 = r3, refs = [], hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st), r0 ≠ r1 ∧ r0 ≠ r2, r3 ≠ r1 ∧ r3 ≠ r2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
(r0 ' + refs1) .last = (r3 ' + refs0) .last, path∃(str, r, r2, st), path∃(str, r, r1, st), validrefs(str, r0 ' + refs1, st[r0 -- _type] .type, st), validrefs(str, r3 ' + refs0, st[r3 -- _type] .type, st) ⊦ r0 = r3, (r3 ' + refs0) .last = jvmref, hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st), r0 ≠ r1 ∧ r0 ≠ r2, r3 ≠ r1 ∧ r3 ≠ r2

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]