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Proof step 11 from proof in-add (previous step)

∀ st, ty, j, r, vals0. # vals0 < # vals1 + 1 → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ r -- j ' ∈ addarray(r, ty, vals1, st) ∨ j = n→i(# vals1) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ n→i(# vals1) < j

User applied rule apply induction

And obtained the following new goals

Proof step 10:
r -- j ' ∈ addarray(r, ty, vals1, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals1) ≤ j ∨ r -- j ' ∈ st, ∀ st, ty, j, r, vals0. # vals0 < # vals1 + 1 → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ r -- j ' ∈ addarray(r, ty, vals1, st) ∨ j = n→i(# vals1) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ n→i(# vals1) < j

Rule argument

[substlist-arg: [st, ty, j, r, vals0] → [st, ty, j, r, vals1]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ ¬ n < n], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (st, ty, j, r, vals1)]]