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Proof step 4 from proof subrec-01 (previous step)

∀ a, y. x0 ⊆m a ' + y ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y ⊦ (a0 = a ∨ a0 ∈ y) ∧ x0 ⊆m (a ' + y) -1l a0 ↔ #oc(a, a0 ') + #oc(a, x0) ≤ #oc(a, y) + 1 ∧ (a0 ' + x0) -1l a ⊆m y, a = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ y, ∀ a, y. x0 ⊆m a ' + y ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y ⊦ x0 ⊆m a ' + (y -1l a0) ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y -1l a0, a = a0

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%subrec: ⊦ a ' + x ⊆m y ↔ a ∈ y ∧ x ⊆m y -1l a;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)