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Proof step 16 from proof addarray-both-subset (previous step)

vals ≠ [], 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 .butlast, st0)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))]) ⊆ addarray(r, byte_type, vals, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
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 .butlast, st0)[r -- n→i(# vals - 1) ', vals .last][r -- _length, intval(n→i(# vals))]) ⊆ addarray(r, byte_type, vals, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n -1 = n - 1])]]