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Proof step 3 from proof app-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ x0, ∀ a, y. a ∈ x0 → pos(a, x0 + y) = pos(a, x0) ⊦ a = a0, pos(a, x0 + y) = pos(a, x0)

Rule argument

Additional information

Used simplifier rules: pos-y: ⊦ pos(a, a ' + x) = 0;%pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list