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Proof step 12 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 ' = r -- n→i(# vals1) ' ∨ r -- j ' ∈ addarray(r, ty, vals1, st) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ n→i(# vals1) < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ 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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ' = i0 ' ↔ i = i0], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0])]]