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Proof step 6 from proof intvalues-at (previous step)

m ≤ # vals0, ∀ m0. ¬ # vals0 ≤ m0 → (intvalues(vals0)[m0]) = vals0[m0] .intval ⊦ m = 0, (intvalues(vals0)[m - 1]) = (val ' + vals0)[m] .intval

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
m = m1 + 1, m ≤ # vals0, ∀ m0. ¬ # vals0 ≤ m0 → (intvalues(vals0)[m0]) = vals0[m0] .intval ⊦ m = 0, (intvalues(vals0)[m1]) = (val ' + vals0)[m] .intval

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [m, m1] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (m, m1)]]