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Proof step 6 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, [], 0, st0[r -- _type, typeval(mkarraytype(byte_type))][r -- _length, intval(n→i(# []))]) ⊆ addarray(r, byte_type, [], st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ addarray(r, [], i, st) = st
Proof step 5:
st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # [] → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ (st0[r -- _type, typeval(mkarraytype(byte_type))][r -- _length, intval(0)]) ⊆ addarray(r, byte_type, [], st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # [] = 0])]]