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Proof step 18 from proof addarray-both-subset (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 8:
vals = [], st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # vals → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ addarray(r, byte_type, vals, st0) ⊆ addarray(r, byte_type, vals, st1)
Proof step 17:
st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # vals → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ vals = [], addarray(r, byte_type, vals, st0) ⊆ addarray(r, byte_type, vals, st1)

Rule argument

[fma-arg: vals = []]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (vals = [])]]