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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.
Let x7 of type ι be given.
Let x8 of type ι be given.
Let x9 of type ι be given.
Let x10 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.
Assume H7: SNo x7.
Assume H8: SNo x8.
Assume H9: SNo x9.
Assume H10: SNo x10.
Apply minus_add_SNo_distr_m_9 with x0, x1, x2, x3, x4, x5, x6, x7, x8, add_SNo (minus_SNo x9) x10, λ x11 x12 . x12 = add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (add_SNo x6 (add_SNo x7 (add_SNo x8 (add_SNo x9 (minus_SNo x10)))))))))) leaving 11 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.
The subproof is completed by applying H5.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
The subproof is completed by applying H8.
Apply SNo_add_SNo with minus_SNo x9, x10 leaving 2 subgoals.
Apply SNo_minus_SNo with x9.
The subproof is completed by applying H9.
The subproof is completed by applying H10.
Apply minus_add_SNo_distr_m with x9, x10, λ x11 x12 . add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (add_SNo x6 (add_SNo x7 (add_SNo x8 x12)))))))) = add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (add_SNo x6 (add_SNo x7 (add_SNo x8 (add_SNo x9 (minus_SNo x10)))))))))) leaving 3 subgoals.
The subproof is completed by applying H9.
The subproof is completed by applying H10.
Let x11 of type ιιο be given.
Assume H11: x11 (add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (add_SNo x6 (add_SNo x7 (add_SNo x8 (add_SNo x9 (minus_SNo x10))))))))))) (add_SNo x0 (add_SNo x1 (add_SNo x2 (add_SNo x3 (add_SNo x4 (add_SNo x5 (add_SNo x6 (add_SNo x7 (add_SNo x8 (add_SNo x9 (minus_SNo x10))))))))))).
The subproof is completed by applying H11.