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Proof step 6 from proof keys-firstn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ fs, ax1. firstn(# ax, keys(fs, ax) + keys(fs, ax1)) = keys(fs, ax) ⊦ firstn(# ax, keys(fs, ax) + keys(fs, ax1)) = keys(fs, ax)

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)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ firstn(n + 1, key ' + kx) = key + firstn(n, kx); (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]% ⊦ m +1 = m + 1; (s)% ⊦ #(adr ' + ax) = (# ax)+1; (s)% ⊦ adr + ax = adr ' + ax; (s)