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Proof step 6 from proof del-del1-nodups (previous step)

∀ a1. x0 -l a1 = x0 -1l a1 ⊦ (a0 ' + x0) -l a = (a0 ' + x0) -1l a, a = a0, dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a1. x0 -l a1 = x0 -1l a1 ⊦ x0 -l a = x0 -1l a, a = a0, dups(x0), a0 ∈ x0

Rule argument

Additional information

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