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Proof step 7 from proof addarray-both-subset (previous step)

st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # [] → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ addarray(r, byte_type, [], st0) ⊆ addarray(r, byte_type, [], st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ addarray(r, ty, vals, st) = addarray(r, vals, 0, st[r -- _type, typeval(mkarraytype(ty))][r -- _length, intval(n→i(# vals))])
Proof step 6:
st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # [] → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ addarray(r, [], 0, st0[r -- _type, typeval(mkarraytype(byte_type))][r -- _length, intval(n→i(# []))]) ⊆ addarray(r, byte_type, [], st1)

Rule argument

[crewrite-arg: addarrayinit-def , F , [st, vals, ty, r] → [st0, [], byte_type, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0, [], byte_type, r)]]