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Proof step 12 from proof addarray-both-subset (previous step)

vals ≠ [], # vals = m + 1, st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < m + 1 → (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(m) ', vals .last][r -- _length, intval(n→i(m) + 1)]) ⊆ (addarray(r, byte_type, vals .butlast, st1)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# vals = m + 1, st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < m + 1 → (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(m) ', vals .last][r -- _length, intval(n→i(m) + 1)]) ⊆ (addarray(r, byte_type, vals .butlast, st1)[r -- n→i(m) ', vals .last][r -- _length, intval(n→i(m) + 1)])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ 0 + m = m], [CSIMPSEQ : ⊦ n ≤ m → m + n0 - n = (m - n) + n0], [CSIMPSEQ : ⊦ n -1 = n - 1])]]