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Proof step 5 from proof diff-wfrec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a, y. x0 \ (a ' + y) = (x0 -l a) \ y, a0 = a ∨ a0 ∈ y ⊦ x0 \ (a ' + y) = ((a0 ' + x0) -l a) \ y

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list