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Proof step 5 from proof del-del1-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a1. x0 -l a1 = x0 -1l a1 ⊦ a = a0, x0 -l a = x0 -1l a, a0 ∈ x0, dups(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);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list