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Proof step 2 from proof keys-length (previous step)

∀ fs. # keys(fs, ax0) = # ax0 ⊦ # keys(fs, adr + ax0) = #(adr + ax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ fs. # keys(fs, ax0) = # ax0 ⊦ # keys(fs, ax0) = # ax0

Rule argument

Additional information

Used simplifier rules: keys-rec: ⊦ keys(fs, adr ' + ax) = fs[adr].key + keys(fs, ax);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ #(adr ' + ax) = (# ax)+1; (s)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ m +1 = m + 1; (s)% ⊦ #(key ' + kx) = (# kx)+1; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ key + kx = key ' + kx; (s)% ⊦ adr + ax = adr ' + ax; (s)