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

# ax0 = n, ax = ax0 + ax1, n ≤ # ax ⊦ firstn(n, keys(fs, ax)) = keys(fs, ax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ firstn(# ax0, keys(fs, ax0) + keys(fs, ax1)) = keys(fs, ax0)

Rule argument

Additional information

Used simplifier rules: keys-append: ⊦ keys(fs, ax0 + ax1) = keys(fs, ax0) + keys(fs, ax1);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ #(ax + ax0) = # ax + # ax0; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]