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Proofgold Proof

pf
Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Let x6 of type ι be given.
Assume H0: SNo x0.
Assume H1: SNo x1.
Assume H2: SNo x2.
Assume H3: SNo x3.
Assume H4: SNo x4.
Assume H5: SNo x5.
Assume H6: SNo x6.
Apply minus_add_SNo_distr_m_5 with x0, x1, x2, x3, x4, add_SNo (minus_SNo x5) x6, λ x7 x8 . x8 = add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (minus_SNo x6)))))) leaving 7 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
The subproof is completed by applying H3.
The subproof is completed by applying H4.
Apply SNo_add_SNo with minus_SNo x5, x6 leaving 2 subgoals.
Apply SNo_minus_SNo with x5.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
Apply minus_add_SNo_distr_m with x5, x6, λ x7 x8 . add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 x8)))) = add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (minus_SNo x6)))))) leaving 3 subgoals.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
Let x7 of type ιιο be given.
Assume H7: x7 (add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (minus_SNo x6))))))) (add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (minus_SNo x6))))))).
The subproof is completed by applying H7.