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Proof step 3 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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
∀ 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

Rule argument

fma: a = a0

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%oc: ⊦ x ⊆m y → ¬ #oc(a, y) < #oc(a, x);% ⊦ (a ' + x) -1l a = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss) Used formulas: a = a0