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Proof step 5 from proof arr2list-get (previous step)

m0 + 1 < # ar, ∀ n, ar. n < # ar ∧ n ≤ m0 ∧ m0 < # ar → (arr2list(ar, m0)[n] ↔ ar[n]) ⊦ (arr2list(ar, m0) + ar[m0 + 1] ')[m0 + 1] ↔ ar[m0 + 1]

User applied rule cut formula

And obtained the following new goals

Proof step 4:
m0 + 1 < # ar, ∀ n, ar. n < # ar ∧ n ≤ m0 ∧ m0 < # ar → (arr2list(ar, m0)[n] ↔ ar[n]) ⊦ # arr2list(ar, m0) = m0 + 1, (arr2list(ar, m0) + ar[m0 + 1] ')[m0 + 1] ↔ ar[m0 + 1]

Rule argument

[fma-arg: # arr2list(ar, m0) = m0 + 1]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ n = # x → ((x + bv ')[n] ↔ bv)], [CSIMPSEQ : ⊦ (x + y) + z = x + y + z])], [EXTRAFMAS : (# arr2list(ar, m0) = m0 + 1)]]