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Proof step 11 from proof addarray-both-subset (previous step)

# 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)])

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [r, st1, st0, vals0] → [r, st1, st0, vals .butlast]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ (st[rk, val][rk0, val0]) ⊆ (st0[rk, val][rk0, val0]) → ¬ st ⊆ st0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ vals ≠ [] → # vals .butlast = # vals - 1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ ¬ m + n < m], [CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : n0 < n ⊦ m + ((* n)) ≤ m0 + ((* n0)) ↔ m + (n - n0) ≤ m0], [CSIMPSEQ : ⊦ m + m0 ≤ m ↔ m0 = 0])], [EXTRATERMS : (r, st1, st0, vals .butlast)]]