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Proof step 10 from proof put-del1-in (previous step)

m0 < # x0, ∀ m, a. m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ∨ x0[m] = a ⊦ a ∈ (a0 ' + x0[m0, a]) -1l a ↔ a = a0 ∨ a ∈ x0, x0[m0] = a, a = a0, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m0 < # x0, ∀ m, a. m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ∨ x0[m] = a ⊦ a ∈ x0[m0, a] -1l a ↔ a ∈ x0, x0[m0] = a, a = a0, x0 = []

Rule argument

Additional information

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