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Proof step 44 from proof hasSharing-pathex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
st[r -- i '] = refval(refs1 .first), 0 ≤ i, 0 < st[r -- _length] .intval, is_array(str, r, st), hasSharing(r0, st), i < st[r -- _length] .intval, validrefs(str, refs1 + r0 ', st[r -- _type] .type .type, st), ∀ st, r, str, refs. # refs < # (refs1 .first ' + refs1 .rest + r0 ') → (validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ hasSharing(refs .last, st) → refs = [] ∨ hasSharing(r, st)) ⊦ r = jvmref, refs1 = [], hasSharing(r, st)

Rule argument

[emptyarg]

Additional information

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