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Proof step 6 from proof RED-complete (previous step)

〈RED1(t'; tn)〉 tn = tn1, ∀ n0, t', t. n0 < 0 → (irred(t') ∧ (=> ^ n0)(t, t') → 〈RED(t; t0)〉 t0 = t'), irred(t'), tn1 = none ⊃ irred(t') ; t' => tn1 .t, ¬ tn1 ≠ none ⊦ 〈t1 := t'〉 t1 = t'

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

[emptyarg]

Additional information

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