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Proof step 3 from proof read-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # ax0, ∀ n, fs. n < # ax0 → fs[ax0][n] = fs[ax0[n]] ∨ ax0 = [] ⊦ (fs[adr] ' + fs[ax0])[n] = fs[(adr ' + ax0)[n]]

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]% ⊦ nd + ndx = nd ' + ndx; (s)% ⊦ fs[adr ' + ax] = fs[adr] + fs[ax]; (s)% ⊦ adr + ax = adr ' + ax; (s)