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Proof step 112 from proof validrefs-lastelem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 111:
acc + r ' = st[r0 -- i '] .refval ' + refs, 0 < st[r0 -- _length] .intval, 0 ≤ i, validrefs(str, acc + r ', st[r0 -- _type] .type .type, st), is_referencevalue(st[r0 -- i ']), i < st[r0 -- _length] .intval, st[r0 -- _type] .type ≤ ty, validrefs(str, r0 ' + acc, ty, st), is_array(str, r0, st), ∀ str, ty, st, r. validrefs(str, acc, ty, st) ∧ validrefs(str, acc + r ', ty, st) → acc = [] ∨ has_reffield(acc .last, st) ⊦ r0 = jvmref, has_reffield((r0 ' + acc) .last, st), r ≠ st[r0 -- i '] .refval ∧ ¬ r ∈ refs, ¬ st[r0 -- i '] .refval ∈ acc ∧ st[r0 -- i '] .refval ≠ r

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])]]