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Proof step 4 from proof keys-at (previous step)

n < #(adr + ax0), ∀ n, fs. n < # ax0 → keys(fs, ax0)[n] = fs[ax0[n]].key ∨ ax0 = [] ⊦ adr + ax0 = [], keys(fs, adr + ax0)[n] = fs[(adr + ax0)[n]].key

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ # ax0, ∀ n, fs. n < # ax0 → keys(fs, ax0)[n] = fs[ax0[n]].key ∨ ax0 = [] ⊦ (fs[adr].key ' + keys(fs, ax0))[n] = fs[(adr ' + ax0)[n]].key

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)% ⊦ #(adr ' + ax) = (# ax)+1; (s)% ⊦ adr ' + ax ≠ []; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ key + kx = key ' + kx; (s)% ⊦ keys(fs, adr ' + ax) = fs[adr].key + keys(fs, ax); (s)% ⊦ adr + ax = adr ' + ax; (s)