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Proof step 2 from proof keys-eq-empty (previous step)

keys(fs, ax) = [] ⊦ ax = []

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: ax

Additional information

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