Implementation [of proof example]

Given the axioms in the form

s[1] = (a_ ⊼ a_) ⊼ (a_ ⊼ b_)  a;
s[2, x_] := b_  (b ⊼ b) ⊼ (b ⊼ x); s[3] = a_ ⊼ (a_ ⊼ b_)  a ⊼ (b ⊼ b); s[4] = a_ ⊼ (b_ ⊼ b_)  a ⊼ (a ⊼ b);
s[5] = a_ ⊼ (a_ ⊼ (b_ ⊼ c_))  b ⊼ (b ⊼ (a ⊼ c));

the proof shown here can be represented by

{{s[2, b], {2}}, {s[4], {}}, {s[2, (b ⊼ b) ⊼ ((a ⊼ a) ⊼ (b ⊼ b))], {2, 2}}, {s[1], {2, 2, 1}}, {s[2, b ⊼ b], {2, 2, 2, 2, 2, 2}], {s[5], {2, 2, 2}}, {s[2, b ⊼ b], {2, 2, 2, 2, 2, 1}}, {s[1], {2, 2, 2, 2, 2}}, {s[3], {2, 2, 2}}, {s[1], {2, 2, 2, 2}}, {s[4], {2, 2, 2}}, {s[5], {}}, {s[2, a], {2, 2, 1}}, {s[1], {2, 2}}, {s[3], {}}, {s[1], {2}}}

and applied using

FoldList[Function[{u, v}, MapAt[Replace[#, v〚1〛] &, u, {v〚2〛}]], a ⊼ b, proof]