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Proof step 122 from proof addarray-put-01 (previous step)

i→n(i) = m + 1, 0 ≤ i0, 0 < i, 0 < i0, i ≤ i0, 0 ≤ i, i→n(i0) ≤ m ⊦ initial_value(ty) = (mklist(initial_value(ty), i→n(i0))[m])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 121:
0 ≤ i ⊦ i→n(i) = ((* m)) ↔ i = n→i(m)

Rule argument

[speclemmaarg: 0 ≤ i ,i→n(i) = ((* m)) ↔ i = n→i(m) ,[m, i] → [m + 1, i] ,F ,intnat , ,num-i2n-eq ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ (~1) < i ↔ 0 ≤ i], [CSIMPSEQ : ⊦ 0 ≤ n→i(n)], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + 1 ≤ j ↔ i < j], [CSIMPSEQ : ⊦ i→n(i) ≤ n → ¬ n→i(n) < i])], [EXTRATERMS : (m + 1, i)]]