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Proof step 12 from proof diff-wfrec (previous step)

∀ a, y. x0 \ (a ' + y) = (x0 -l a) \ y ⊦ (a0 ' + x0) \ (a ' + y) = ((a0 ' + x0) -l a) \ y, a0 ∈ a ' + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
∀ a, y. x0 \ (a ' + y) = (x0 -l a) \ y ⊦ x0 \ (a ' + y) = (x0 -l a) \ y, a0 = a, a0 ∈ y

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a); (s)% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)