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Proof step 31 from proof cont-monotone (
previous step
)
(n = 0 ⊃ S0;S)(x0), f(S0)(x), S0 ⊆ S, f(λ x0. ∃ n. (n = 0 ⊃ S0;S)(x0))(x) ⊦ S(x0), f(S)(x)
Heuristic if-then-else split applied rule case distinction
And obtained the following new goals
Proof step 32
:
n = 0, S0(x0), f(S0)(x), S0 ⊆ S, f(λ x0. ∃ n. (n = 0 ⊃ S0;S)(x0))(x) ⊦ S(x0), f(S)(x)
Rule argument
pos: left 1
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