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Proof step 2 from proof len-01 (previous step)

# x < pos(a, x) ⊦ pos(a, x) = 0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a. # x0 < pos(a, x0) → pos(a, x0) = 0, #(a0 + x0) < pos(a, a0 + x0) ⊦ pos(a, a0 + x0) = 0

Rule argument

term: x

Additional information

Used simplifier rules: pos-e: ⊦ pos(a, []) = 0;%list freely generated by [], +;;