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Proof step 28 from proof switch-array (previous step)

∀ st, vals0, ty0, ty, r0, r, vals1. # vals1 < # vals → (true → r = r0 ∨ addarray(r, ty, vals1, addarray(r0, ty0, vals0, st)) = addarray(r0, ty0, vals0, addarray(r, ty, vals1, st))) ⊦ r = r0, vals = [], (addarray(r, ty, vals .butlast, addarray(r0, ty0, vals0, st))[r -- n→i(# vals - 1) ', vals .last][r -- _length, intval(n→i(# vals))]) = addarray(r0, ty0, vals0, addarray(r, ty, vals, st))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 27:
# vals = m + 1 ∧ # vals ≠ 0, ∀ st, vals0, ty0, ty, r0, r, vals1. # vals1 < # vals → (true → r = r0 ∨ addarray(r, ty, vals1, addarray(r0, ty0, vals0, st)) = addarray(r0, ty0, vals0, addarray(r, ty, vals1, st))) ⊦ r = r0, vals = [], (addarray(r, ty, vals .butlast, addarray(r0, ty0, vals0, st))[r -- n→i(m) ', vals .last][r -- _length, intval(n→i(# vals))]) = addarray(r0, ty0, vals0, addarray(r, ty, vals, st))

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [# vals, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # vals = 0 ↔ vals = []], [CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (# vals, m)]]