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Proof step 50 from proof hasSharing-pathex (previous step)

refs1 + r0 ' = r1 + refs0 ∧ refs1 + r0 ' ≠ [], r1 = st[r -- i '] .refval, is_array(str, r, st), hasSharing(r0, st), i < st[r -- _length] .intval, is_referencevalue(st[r -- i ']), validrefs(str, refs1 + r0 ', st[r -- _type] .type .type, st), 0 ≤ i, 0 < st[r -- _length] .intval, ∀ st, r, str, refs. # refs < # (refs1 + r0 ') → (validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ hasSharing(refs .last, st) → refs = [] ∨ hasSharing(r, st)) ⊦ r = jvmref, hasSharing(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 49:
refs1 + r0 ' = st[r -- i '] .refval ' + refs0, 0 < st[r -- _length] .intval, 0 ≤ i, validrefs(str, refs1 + r0 ', st[r -- _type] .type .type, st), is_referencevalue(st[r -- i ']), i < st[r -- _length] .intval, hasSharing(r0, st), is_array(str, r, st), ∀ st, r, str, refs. # refs < # (refs1 + r0 ') → (validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ hasSharing(refs .last, st) → refs = [] ∨ hasSharing(r, st)) ⊦ r = jvmref, hasSharing(r, st), r0 ≠ st[r -- i '] .refval ∧ ¬ r0 ∈ refs0, ¬ st[r -- i '] .refval ∈ refs1 ∧ st[r -- i '] .refval ≠ r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0], [CSIMPFORWARD : ⊦ refs + r ' = refs0 → r ∈ refs0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ refs + r ' ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]