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Proof step 3 from proof write-shift (previous step)

# ax = # ndx ⊦ fs[adr, nd][ax, ndx] = fs[ax, ndx][adr, nd], adr ∈ ax

User applied rule structural induction

And obtained the following new goals

Proof step 4:
#(adr0 + ax0) = # ndx, ∀ ndx, fs, adr, nd. # ax0 = # ndx → fs[adr, nd][ax0, ndx] = fs[ax0, ndx][adr, nd] ∨ adr ∈ ax0 ⊦ fs[adr, nd][adr0 + ax0, ndx] = fs[adr0 + ax0, ndx][adr, nd], adr ∈ adr0 + ax0

Rule argument

indvar: ax

Additional information

Used simplifier rules: write-base: ⊦ fs[[], []] = fs;%address-list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ # ndx = 0 ↔ ndx = []; (s)% ⊦ ¬ adr ∈ []; (s)