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Proof step 40 from proof reffield-del (previous step)

is_array(r0, st), has_reffield(r0, st) ⊦ r0 = r, has_reffield(r0, st -- r)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ is_array(r, st) → (has_reffield(r, st) ↔ st[r -- _length] .intval > 0 ∧ (is_reftype(st[r -- _type] .type .type) ∨ st[r -- _type] .type .type = void_type))
Proof step 39:
is_array(r0, st), is_array(r0, st), st[r0 -- _length] .intval > 0 ∧ (is_reftype(st[r0 -- _type] .type .type) ∨ st[r0 -- _type] .type .type = void_type) ⊦ r0 = r, has_reffield(r0, st -- r)

Rule argument

[crewrite-arg: has_reffield-array, F, [st, r] → [st, r0]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r0)]]