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Proof step 11 from proof RED-complete (previous step)

∀ n0, t', t. n0 < 0 → (irred(t') ∧ (=> ^ n0)(t, t') → 〈RED(t; t0)〉 t0 = t'), irred(t') ⊦ 〈let tn = none in { RED1(t'; tn) ; if tn ≠ none then RED(tn .t; t1) else t1 := t' }〉 t1 = t'

Heuristic symbolic execution applied rule vardecls right

And obtained the following new goals

Proof step 10:
∀ n0, t', t. n0 < 0 → (irred(t') ∧ (=> ^ n0)(t, t') → 〈RED(t; t0)〉 t0 = t'), irred(t'), tn0 = none ⊦ 〈RED1(t'; tn0)〉〈if tn0 ≠ none then RED(tn0 .t; t1) else t1 := t'〉 t1 = t'

Rule argument

[emptyarg]

Additional information

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