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

∀ fs, ax1. keys(fs, ax + ax1) = keys(fs, ax) + keys(fs, ax1) ⊦ keys(fs, (adr + ax) + ax1) = keys(fs, adr + ax) + keys(fs, ax1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ fs, ax1. keys(fs, ax + ax1) = keys(fs, ax) + keys(fs, ax1) ⊦ keys(fs, ax + ax1) = keys(fs, ax) + keys(fs, ax1)

Rule argument

Additional information

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