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Proof step 6 from proof append-to-le-01 (previous step)

m0 ≤ # x0, ∀ x0, a, σ. m0 ≤ # x0 + 1 ∧ (∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ to m = ⌜x0 to m⌝ ) → (⌜a '⌝ + ⌜x0⌝ ) + σ to m0 = ⌜a ' + x0 to m0⌝ , ∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ to m = ⌜x0 to m⌝ ⊦ ⌜x0⌝ + σ to m0 = ⌜x0 to m0⌝

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: m0, σ

Additional information

Used terms: m0, σ