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Proof step 4 from proof app-05-neg (previous step)

∀ a, y. pos(a, x0 + y) = pos(a, y) + # x0 ∨ a ∈ x0 ⊦ pos(a, (a0 + x0) + y) = pos(a, y) + #(a0 + x0), a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a, y. pos(a, x0 + y) = pos(a, y) + # x0 ∨ a ∈ x0 ⊦ a = a0, pos(a, x0 + y) = pos(a, y) + # x0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)