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Proof step 4 from proof app (previous step)

a ∈ a0 + x0, ∀ a, y. a ∈ x0 → (x0 + y) -1l a = (x0 -1l a) + y ⊦ ((a0 + x0) + y) -1l a = ((a0 + x0) -1l a) + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x0, ∀ a, y. a ∈ x0 → (x0 + y) -1l a = (x0 -1l a) + y ⊦ (x0 + y) -1l a = (x0 -1l a) + y, a = a0

Rule argument

Additional information

Used simplifier rules: del1-y: ⊦ (a ' + x) -1l a = x;%del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)